objects 1 and 2 are connected by a tight rope. object 1 moves with an acceleration magnitude a 1, and object 2 moves with an acceleration magnitude a 2. which is true?

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Answer 1

When objects 1 and 2 are connected by a tight rope, their accelerations are interrelated. If the rope remains taut and the system is considered in isolation, then the net force acting on the entire system must be equal. Consequently, the accelerations of both objects (a1 and a2) are determined by the ratio of their respective masses (m1 and m2) and the external forces acting on them.

In this scenario, if a larger force is acting on object 1 compared to object 2, the acceleration magnitude a1 will be greater than a2. Conversely, if a larger force is acting on object 2, the acceleration magnitude a2 will be greater than a1. However, if the external forces acting on both objects are equal, then the acceleration magnitudes a1 and a2 will also be equal.

To summarize, the relationship between the acceleration magnitudes a1 and a2 depends on the external forces acting on the objects and their respective masses. By analyzing the forces and mass ratios, one can determine which object will have a greater acceleration magnitude or if they will be equal.

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write an expression for the component of net force, fnet,x, in the x-direction, in terms of the variables given in the problem statement.

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The component of the net force, fnet,x, in the x-direction can be expressed as the sum of all forces acting in that direction: fnet,x = ΣFx.

To find the component of net force, fnet,x, in the x-direction, we need to consider all the forces acting in that direction. Let's assume there are n forces acting in the x-direction. Then we can write:

ΣFx = F1x + F2x + F3x + ... + Fnx

where F1x, F2x, F3x, ..., Fnx are the x-components of the individual forces.

We can then substitute the expressions for each of the individual x-components of the forces. For example, if we have a force F1 acting at an angle θ1 to the x-axis, we can use trigonometry to find its x-component:

F1x = F1 cos(θ1)

Similarly, for all the other forces, we can find their x-components and add them up to get the total sum of forces in the x-direction. This gives us the expression for the component of the net force, fnet,x, in the x-direction:

fnet,x = ΣFx

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A 12-V battery causes a current of 0.80 A through a resistor.
a) What is its resistance?
b) How many joules of energy does the battery lose in a minute?
2) You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240 V. If you use the lightbulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 120-V bulbs?

Answers

a) The resistance will be:

R = V / I = 12 V / 0.80 A = 15 Ω

b) E = P * t = 9.6 W * 60 s = 576 Joules

2) The 75 W lightbulb used at 120 V will have a relative brightness of 18.75 W 120 V bulbs.

How to find resistance?

a) To calculate the resistance, we can use Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I). Therefore, the resistance can be calculated as follows:

R = V / I = 12 V / 0.80 A = 15 Ω

How to find joules of energy of battery lose in a minute?

b) The power (P) consumed by the battery can be calculated using the formula: P = V * I, where V is the voltage and I is the current. The energy (E) consumed in a given time period can be calculated by multiplying power (P) by time (t). In this case, since we want to find the energy consumed in a minute, the time is 60 seconds.

P = V * I = 12 V * 0.80 A = 9.6 W

E = P * t = 9.6 W * 60 s = 576 Joules

How to find brightness of a bulb?

The power of the lightbulb remains constant regardless of the voltage applied, so the power rating of the lightbulb is still 75 W. However, the brightness of a lightbulb is typically measured in terms of its luminous flux, which is not directly proportional to power. Luminous flux is measured in lumens (lm).

To determine how bright the 75 W lightbulb will be at 120 V relative to 75 W 120 V bulbs, we need to compare the luminous flux. Assuming the lightbulb's resistance remains constant, we can use the formula for power (P) in terms of resistance (R) and voltage (V): P = V^2 / R.

For the 75 W lightbulb at 240 V:

P1 = 75 W

V1 = 240 V

For the lightbulb at 120 V:

P2 = ?

V2 = 120 V

Using the formula, we can solve for P2:

P1 / P2 = (V1 / V2)^2

75 W / P2 = (240 V / 120 V)^2

75 W / P2 = 2^2

75 W / P2 = 4

P2 = 75 W / 4 = 18.75 W

Therefore, the 75 W lightbulb used at 120 V will have a relative brightness of 18.75 W 120 V bulbs.

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the lowest three resonant frequencies that can be produced in a hollow tube are as follows:200 hz, 600 hz, 1000hzwhat kind of tube is it?

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The three resonant frequencies provided are in a 1:3:5 ratio, which is characteristic of open-ended tubes. The fundamental frequency (first harmonic) is 200 Hz, while the second and third harmonics are 600 Hz and 1000 Hz, respectively. These harmonics indicate the presence of an open-ended tube.

The tube in question is likely a closed cylindrical tube with one end closed and one end open. This is based on the fact that the lowest resonant frequency of a closed cylindrical tube is approximately 200 Hz, while the second and third resonant frequencies are approximately three times and five times higher, respectively.  It's important to understand what resonant frequencies are and how they are produced in a hollow tube. Resonant frequencies are the natural frequencies at which an object vibrates when it is excited by an external force.

The resonant frequencies of a hollow tube depend on the length and shape of the tube, as well as the speed of sound in the medium inside the tube (usually air). The resonant frequencies of a closed cylindrical tube are given by the formula f(n) = n*v/2L, where f(n) is the frequency of the nth resonant mode, v is the speed of sound, L is the length of the tube, and n is an integer representing the number of half-wavelengths that fit inside the tube. For a closed cylindrical tube with one end closed and one end open, the lowest resonant frequency (n=1) is approximately 200 Hz, while the second (n=3) and third (n=5) resonant frequencies are approximately 600 Hz and 1000 Hz, respectively. Based on the given resonant frequencies (200 Hz, 600 Hz, and 1000 Hz), it seems that you are dealing with an open-ended tube

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A car initially at rest accelerates at 10 m/s^2. The car's speed after it has traveled 25 meters is most nearly

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So, the car's speed after it has traveled 25 meters is most nearly 22.36 m/s.

The equation for acceleration is a = (v_f - v_i) / t, where a is acceleration, v_f is final velocity, v_i is initial velocity (in this case, zero), and t is time. We know that a = 10 m/s^2 and we want to find v_f after the car has traveled 25 meters. So we need to rearrange the equation to solve for v_f:

a = (v_f - v_i) / t
10 m/s^2 = (v_f - 0) / t
10 m/s^2 * t = v_f

Now we need to find t. We can use the equation d = v_i * t + 1/2 * a * t^2, where d is distance. We know that d = 25 meters, v_i = 0, and a = 10 m/s^2, so we can plug those values in and solve for t:

25 meters = 0 * t + 1/2 * 10 m/s^2 * t^2
25 meters = 5t^2
t^2 = 5 meters
t = sqrt(5) meters (since t has to be positive)

Now we can plug in t to find v_f:

v_f = 10 m/s^2 * sqrt(5) meters
v_f = 22.36 m/s (rounded to two decimal places)

So the car's speed after it has traveled 25 meters is most nearly 22.36 m/s.
A car initially at rest accelerates at 10 m/s² and travels 25 meters. To find its speed after traveling this distance, we can use the equation:

v² = u² + 2as

where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration (10 m/s²), and s is the distance traveled (25 m).

v² = 0² + 2(10)(25)
v² = 0 + 500
v² = 500

Now, we'll take the square root of both sides to find the final velocity:

v = √500
v ≈ 22.36 m/s

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in an 8.00 km race, one runner runs at a steady 11.1 km/h and another runs at 14.1 km/h. how far from the finish line is the slower runner when the faster runner finishes the race?

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when the faster runner finishes the race, the slower runner still has 8.00 km - 6.30 km = 1.70 km left to run before reaching the finish line.

To find out how far from the finish line the slower runner is when the faster runner finishes the race, we need to first calculate how long it takes for the faster runner to finish the race.
Using the formula:
time = distance ÷ speed
We can calculate the time it takes for the faster runner to finish the race:
time = 8.00 km ÷ 14.1 km/h
time = 0.567 hours
Now we know that the faster runner finishes the race in 0.567 hours.
To find out how far the slower runner is from the finish line when the faster runner finishes the race, we need to calculate how far the slower runner has run in the same amount of time (0.567 hours).
distance = speed x time
For the slower runner:
distance = 11.1 km/h x 0.567 hours
distance = 6.30 km
Therefore, when the faster runner finishes the race, the slower runner still has 8.00 km - 6.30 km = 1.70 km left to run before reaching the finish line.
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At a given rotational speed, how does linear (or tangential) speed change as the distance from the axis changes?

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At a given rotational speed, linear (or tangential) speed increases as the distance from the axis increases.

This relationship is described by the equation:

v = rω

where v is the tangential velocity, r is the distance from the axis (i.e., the radius), and ω is the angular velocity.

This means that for a given rotational speed (i.e., angular velocity), objects farther from the axis will be moving faster than objects closer to the axis. This relationship is demonstrated in everyday examples such as the rotation of a bicycle wheel, where the speed of the outer edge is much greater than the speed of the hub.

It's important to note that this relationship assumes a constant angular velocity. If the angular velocity changes, then the linear speed will also change, regardless of the distance from the axis.

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The temperature of an example of CH4 gas (10.34g) in a 50.0L vessel at 1.33atm is___
A. 984
B. -195
C. 195
D. 1260
E. -1260

Answers

The temperature of the [tex]CH_4[/tex] gas is 195 K (Option C).

We can use the ideal gas law to solve this problem: PV = nRT

where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in K.

First, we need to calculate the number of moles of [tex]CH_4[/tex] using its molar mass:

Molar mass of [tex]CH_4[/tex] = 12.01 g/mol (C) + 4(1.01 g/mol) = 16.05 g/mol

Number of moles of [tex]CH_4[/tex] = 10.34 g / 16.05 g/mol = 0.644 mol

Now, we can rearrange the ideal gas law to solve for T:

T = PV / (nR)

Substituting the given values, we get:

T = (1.33 atm) x (50.0 L) / (0.644 mol x 0.0821 L·atm/mol·K) = 195 K

Therefore, the temperature of the [tex]CH_4[/tex] gas is 195 K (Option C).

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type ia and type ii supernovae are respectively caused by what types of stars?

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Both Type Ia and Type II supernovae are important astronomical events that provide us with valuable information about the life cycle of stars and the formation of the universe.


Type Ia and Type II supernovae are two different types of supernovae that are caused by different types of stars. Type Ia supernovae are caused by white dwarf stars, which are the remnants of stars that have exhausted all of their nuclear fuel and have collapsed to a very small size. These stars are typically in a binary system with another star, and they can accrete matter from their companion star. When a white dwarf reaches a certain mass, it can undergo a runaway nuclear reaction that causes it to explode as a supernova.
On the other hand, Type II supernovae are caused by much more massive stars, which have exhausted their nuclear fuel and can no longer support their own weight. These stars undergo a series of complex nuclear reactions that result in the production of heavier elements, and eventually, they collapse under their own gravity and explode as supernovae.
Overall, both Type Ia and Type II supernovae are important astronomical events that provide us with valuable information about the life cycle of stars and the formation of the universe. By studying these explosions and their remnants, astronomers can learn more about the composition and evolution of the universe, as well as the origins of the chemical elements that make up our world.

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Molecular spectra, like elemental one, involve only the vibration of the particles. ture or false?

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False. While elemental spectra typically involve the emission or absorption of light due to electronic transitions within an atom, molecular spectra involve the vibration and rotation of the constituent atoms within a molecule.

Molecules have more degrees of freedom than atoms, which leads to more complex spectra. In addition to electronic transitions, the energy levels of molecules are also affected by their vibrational and rotational motion. When a molecule absorbs or emits light, it can undergo changes in both its electronic and vibrational/rotational states, leading to a more complex spectrum.

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An 80-km/h airplane caught in a 60-km/h crosswind has a resultant speed of
Select one:
a. 141 km/h.
b. 60 km/h.
c. 100 km/h.
d. 80 km/h.

Answers

The answer to this question is a. 141 km/h. The airplane's speed and the crosswind speed are not added together to get the resultant speed because they are not in the same direction. Instead, we use the Pythagorean theorem to calculate the resultant speed.

To find the resultant speed, we need to use the Pythagorean theorem because the airplane's speed and the crosswind speed are perpendicular to each other. The Pythagorean theorem states that the square of the hypotenuse (resultant speed) is equal to the sum of the squares of the other two sides (airplane speed and crosswind speed). Using this formula, we can calculate the resultant speed as follows:
Resultant speed = √(80^2 + 60^2)
Resultant speed = √(6400 + 3600)
Resultant speed = √10000
Resultant speed = 100 km/h
Therefore, the answer is not d. 80 km/h, but rather a. 141 km/h.

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What is the relationship between wavelength and frequency quizlet.

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Answer:

Wavelength is inversely proportional to the frequency.

Explanation:

V=fλ, where V is the speed of wave , f is the frequency of wave and lamda is the Wavelength of wave.

So rearranging the above formula : λ=v/f so λ∝ 1/f

Objects that are not actively moving but have the capacity to do so are said to possess:
a. kinetic energy.
b. entropy potential energy.
c. living energy.

Answers

Objects that are not actively moving but have the capacity to do so possess a.) kinetic energy.

So, the correct answer is a. kinetic energy.

Kinetic energy is the energy that an object possesses due to its motion. Even if an object is not currently in motion, if it has the capacity to move, it possesses potential kinetic energy. This energy can be released when the object is set into motion. Entropy potential energy and living energy are not relevant in this context. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.

So, Objects that are not actively moving but have the capacity to do so possess a.) kinetic energy.

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the fairly flat, circular part of the galaxy is referred to as the _______.

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The fairly flat, circular part of a galaxy is referred to as the "galactic disk" or the "stellar disk."

The galactic disk is one of the main components of a spiral galaxy and contains the majority of a galaxy's stars, as well as various interstellar materials like gas and dust.

The disk has a flattened shape, with stars and other objects orbiting the galaxy's central bulge. It is within the galactic disk that most of the star formation and ongoing stellar activity occur.

The galactic disk is often characterized by spiral arms, where regions of higher star density and star formation are observed.

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compared to earth, the moon lacks a hydrosphere, atmosphere, and a magnetosphere. true or false

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True

Compared to Earth, the Moon lacks a hydrosphere (a system of water on its surface, including oceans, lakes, and rivers), an atmosphere and a magnetosphere (a region of space influenced by a planet's magnetic field).

The Moon's surface is dry and lacks significant amounts of water, it has an extremely thin or negligible atmosphere, and it has no substantial magnetic field to create a magnetosphere.

Here's a more detailed explanation:

Hydrosphere: The hydrosphere refers to the presence of water in liquid form on a planetary body. Earth has a significant hydrosphere, with about 71% of its surface covered by water in the form of oceans, seas, lakes, and rivers.

In contrast, the Moon lacks a substantial hydrosphere. While there is evidence of water ice in permanently shadowed regions near the Moon's poles, it is in the form of solid ice rather than liquid water.

Atmosphere: Earth has a dense atmosphere composed primarily of nitrogen (about 78%) and oxygen (about 21%), along with other trace gases. The atmosphere plays a crucial role in regulating temperature, supporting life, and protecting the planet from harmful radiation.

In contrast, the Moon has an extremely thin and tenuous atmosphere, often referred to as an exosphere. It consists of extremely low-density particles, such as atoms and ions, and is practically nonexistent compared to Earth's atmosphere.

Magnetosphere: Earth has a magnetic field generated by its liquid iron outer core. This magnetic field extends into space and creates a region around the planet known as the magnetosphere.

The magnetosphere protects Earth from the solar wind, a stream of charged particles emitted by the Sun. The Moon, however, lacks a significant magnetic field.

It does not have a liquid iron core like Earth, and thus, it does not generate a magnetosphere. As a result, the Moon is directly exposed to the solar wind and its associated radiation.

The absence of a hydrosphere, atmosphere, and magnetosphere on the Moon significantly influences its surface conditions and overall environment.

These factors contribute to the Moon's starkly different appearance and characteristics compared to Earth.

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When a hybrid car brakes to a stop much of its kinetic energy is transformed to a) heat b) work c) electric potential energy

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When a hybrid car brakes to a stop, much of its kinetic energy is transformed into different forms of energy. One of the primary forms of energy that is produced during this process is heat. As the brakes of the car are applied, the friction between the brake pads and the wheels creates heat. This heat is then dissipated into the air around the car, resulting in a loss of energy.



Hybrid cars are designed to capture some of this lost energy and convert it into useful forms of energy that can be used to power the car. In many cases, the kinetic energy that is lost during braking is converted into electric potential energy, which is then stored in the car's battery. This energy can then be used to power the car's electric motor, which in turn can help reduce the car's overall fuel consumption.


It is through the conversion of kinetic energy into electric potential energy or the conversion of energy into work, hybrid cars are a great example of how technology can be used to improve the efficiency of vehicles and reduce their environmental impact.

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you place an object a distance of 30 cm in front of a lens with a focal length of -20 cm. where will the image be formed (in cm) ?

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The image will be formed 60 cm behind the lens, on the same side as the object, but it will be a virtual image, which means that it cannot be projected onto a screen.

To determine the location of the image formed by a lens, we can use the thin lens equation:

1/f = 1/di + 1/do

where f is the focal length of the lens, di is the distance from the lens to the image, and do is the distance from the lens to the object. We can rearrange this equation to solve for di:

1/di = 1/f - 1/do

Plugging in the values given in the problem, we have:

1/-20 = 1/di + 1/30

Simplifying and solving for di, we get:

di = -60 cm

The negative sign for the image distance indicates that the image is formed on the opposite side of the lens from the object, which means it is a virtual image.

Therefore, the image will be formed 60 cm behind the lens, on the same side as the object, but it will be a virtual image, which means that it cannot be projected onto a screen.

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a single slit is the simplest means to produce an interference pattern from light waves. how does the pattern of light and dark bands change as the slit gets wider?

Answers

Answer:

One must consider the phase difference between each side of the slit:

One can write Δ = W sin θ       where Δ is the phase difference between light leaving opposite sides of the slit, W = slit width, and θ is the angle of difraction

As the slit gets wider (W increases) and for a  given phase difference the angle of difraction will decrease

The bands on the screen will be closer for a wider slit

a sprinter runs 100.0 m in 9.87 seconds. if he travels at constant acceleration for the first 75.0 m and then at constant velocity for the final 25.0 m, what was his acceleration during the first 75.0 m?

Answers

The acceleration of the sprinter during the first 75.0 m was 1.44 m/[tex]s^2.[/tex]

We can use the kinematic equations of motion to solve this problem. Let's assume that the sprinter has an initial velocity of zero at the starting point, and a final velocity of v at the end of the 75.0 m distance. We can also assume that the time taken to cover the first 75.0 m is [tex]t_1,[/tex] and the time taken to cover the last 25.0 m is [tex]t_2[/tex].

For the first 75.0 m, we can use the following kinematic equation:

[tex]d = (1/2)at1^2[/tex]

where d is the distance covered, a is the acceleration, and t1 is the time taken to cover the distance.

For the last 25.0 m, we can use the following kinematic equation:

[tex]d = vt_2[/tex]

where d is the distance covered, v is the final velocity, and [tex]t_2[/tex] is the time taken to cover the distance.

We can also use the following kinematic equation for the entire 100.0 m distance:

[tex]d = (1/2)at^2[/tex]

where d is the distance covered, a is the acceleration, and t is the total time taken to cover the distance.

Using the given values of distance and time, we can write the following three equations:

75.0 m = [tex](1/2)at1^2[/tex] (equation 1)

25.0 m =[tex]vt_2[/tex] (equation 2)

100.0 m = [tex](1/2)at^2[/tex] (equation 3)

Since the sprinter covers the last 25.0 m at constant velocity, we know that his final velocity, v, is the same as his average velocity over the last 25.0 m. Therefore, we can write:

v = 25.0 m / t2

Substituting this expression for v into equation 3, we get:

100.0 m = [tex](1/2)at1^2[/tex] + 25.0 m / [tex]t_2[/tex]

Simplifying this equation, we get:

200.0 m = at1^2 + 50.0 m / [tex]t_2[/tex]

Now we can use equation 1 to eliminate t1:

[tex]t_1 =\sqrt(2d/a)[/tex]

Substituting this expression for t1 into equation 2, we get:

25.0 m = [tex]v(\sqrt(2d/a))[/tex]

Simplifying this equation, we get:

[tex]v^2[/tex]= 50.0ad

Substituting this expression for [tex]v^2[/tex] into the previous equation, we get:

200.0 m = (a/2)(2d/a) + (d/a) [tex]v^2[/tex]

Simplifying this equation, we get:

200.0 m = d(1/2 + 1/2)

or

d = 200.0 m

Substituting this value of d into equation 3, we get:

200.0 m = [tex](1/2)at^2[/tex]

Simplifying this equation, we get:

a =[tex](2d/t^2)[/tex]

Substituting the given values of distance and time, we get:

a = (2 x 75.0 m / (9.87 [tex]s)^2)[/tex]

a = 1.44 m[tex]/s^2[/tex]

Therefore, the acceleration of the sprinter during the first 75.0 m was 1.44 m/[tex]s^2.[/tex]

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a biconcave (diverging) lens has equal radii of curvature of 15.1 cm. an object placed 14.2 cm from the lens forms a virtual image 5.29 cm from the lens. what is the index of refraction of the lens material? answer: 1.90

Answers

The index of refraction of the lens material is 2.68.

The refractive index of a lens material is given by the formula; `n = h/h'`.

Where h is the object height and h' is the height of the image.An image is created in the diverging lens that is biconcave. This means that the lens would not cause the light rays to converge but would cause them to diverge instead.

This indicates that the focal length of the lens will be negative.

As such, the lens equation to use will be `1/f = 1/v - 1/u`.

The negative sign will be assigned to the focal length of the lens to represent the fact that it is a diverging lens.

The values are:f = -15.1cm; u = -14.2cm; v = 5.29cm.

We have: `1/f = 1/v - 1/u = 1/5.29 + 1/14.2`.

Solving for f we get: `f = -9.59 cm`.

Then, the magnification `m = -v/u = 5.29/14.2 = 0.373`.

The object height can be calculated using the formula; `m = h'/h`.

From this; `h' = mh`. `h = -h' / m = -5.29 / 0.373 = -14.17 cm`.

Since the height is negative, it indicates that the object is inverted.

The refractive index `n = h/h' = 14.17/5.29 = 2.68`.

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a major-league pitcher can throw a baseball at 41 m/sec. if a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17 m away from the point of release?

Answers

This means the ball drops about 0.87 meters (about 2 feet, 10 inches) by the time it reaches the catcher.

To answer this question, we need to use the equation for projectile motion, which is:
y = yo + voyt + 1/2at^2

where y is the vertical displacement, yo is the initial vertical position, voy is the initial vertical velocity, t is time, and a is the acceleration due to gravity.

In this case, the ball is thrown horizontally, so there is no initial vertical velocity (voy = 0), and the initial vertical position is also zero (yo = 0). We know the distance the ball travels horizontally (17 m) and the initial speed (41 m/s), so we can use the equation:
x = vot + 1/2at^2

where x is the horizontal displacement and vo is the initial horizontal velocity. Solving for t, we get:
t = x / vo = 17 / 41 = 0.4146 s

Now we can use the equation for vertical displacement to find how much the ball drops during that time. Since we know the acceleration due to gravity is -9.8 m/s^2 (downward), we get:
y = 1/2at^2 = 1/2(-9.8)(0.4146)^2 = -0.8735 m

This means the ball drops about 0.87 meters (about 2 feet, 10 inches) by the time it reaches the catcher. It's important to note that this calculation assumes the ball is thrown perfectly horizontally, with no vertical component to its motion. In reality, a pitched ball will have some degree of arc, which will affect its trajectory and how much it drops before reaching the catcher.

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a student put the charged polythene rod onto a balance. The rod was seperated from the metal pan of the balance by a thin block of insulating material. The student then held a second charged polythene rod above, but not touching, the first rod. The reading on the balance increased

Answers

The phenomenon which made the first polythene rod heavier is because the electric charges on it made it attract more air molecules thus making it heavier.

What is a polythene rod used for?

A charged polythene rod can be used to transmit charge to an insulated conductor without contacting the two objects. The conductor is approached by bringing the negatively charged polythene rod close to it.

Electrons generate enough energy to 'rub off' onto the polythene rod and depart the atom. Electrons are rubbed off the acetate and onto the duster when the rod is replaced with a new substance, such as acetate. The rods and duster are both comprised of insulating materials.

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Full Question:

A student pit the charged polythene rod on to a balance.  The rod was separated from the metal pan of the balance by a thin block of ons laying material.  The student then held a second charged polythene rod above,  but not touching the rod.  The reading on the balance increased.  Explain

The position of a small object is given by x = 34 + 10 − 2t2, where is in seconds and x in meters.
(a) Plot x as a function of t from t = 0 to t = 3.0 s. (b) Find the average velocity of the object between 0 and 3.0 s. (c) At what time between 0 and 3.0 s is the instantaneous velocity zero?

Answers

The instantaneous velocity is zero at t = 0 seconds.(a)

To plot x as a function of t, we can substitute values of t from 0 to 3.0 seconds into the equation x = 34 + 10 − 2t2. This gives us the following values of x: at t=0, x=34+10=44; at t=1.0, x=34+10−2(1.0)2=42; at t=2.0, x=34+10−2(2.0)2=30; at t=3.0, x=34+10−2(3.0)2=10. We can then plot these values on a graph to get the graph of x as a function of t.

(b) The average velocity of the object between 0 and 3.0 seconds can be found by dividing the change in position (x) by the change in time (t). The change in position is x(3.0) − x(0) = 10 − 44 = −34 meters, and the change in time is 3.0 − 0 = 3.0 seconds. Therefore, the average velocity is −34/3 = −11.3 m/s.

(c) The instantaneous velocity is given by the derivative of the position function, which is dx/dt = −4t. The instantaneous velocity is zero when −4t = 0, or when t = 0. Therefore, the instantaneous velocity is zero at t = 0 seconds.

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Hey guys….. anyone knows how to do this?

Answers

Answer:

D

Explanation:

A column of alcohol will be longer by a factor which is the ratios of the densities 13600/789.  

The length of the mercury column replaced by the alcohol is 75.6 - 73.2 cm.

Hence h = (75.6-73.2) x 13600/789 = 41.4 cm

Where would the weight of an object be the least?

Where would the weight of an object be the least?



clear 1. At the equator

2. 500 miles above Earth's surface.

3. At the North pole

4. At the South pole.

5. On the Moon.

Answers

Answer: 5. On the Moon.

Explanation: Weight is a measure of the force of gravity acting on an object.  The weight of an object depends on the mass of the object and the strength of the gravitational force at a particular location.

On Earth, the weight of an object is determined by the mass of the object and the strength of Earth's gravitational force. At the equator, the weight of an object is slightly less compared to the poles due to the centrifugal force caused by the Earth's rotation. This force counteracts a small portion of the gravitational force, resulting in a slightly lower weight.

At the North and South poles, the weight of an object is slightly higher compared to the equator due to the shape of the Earth. The Earth is not a perfect sphere but slightly flattened at the poles, which causes objects at the poles to be closer to the center of the Earth and experience a slightly stronger gravitational force.

However, on the Moon, the weight of an object is significantly less compared to Earth.  The Moon has a much smaller mass and weaker gravitational force than Earth, resulting in objects weighing less on the lunar surface.

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Find the magnitude of the sum
of these two vectors:
B
3.14 m
2.71 m
30.0°
-60.0°

Answers

The magnitude of the sum of two vectors A and B is 4.13 m and the angle of the resultant vector is 10.86°.

From the given,

A = 3.14 m

B = 2.71 m

The resultant vector C= A + B

Vector A is resolved into its vertical and horizontal components,

Aₓ = 3.14 cos(30) = 2.71 m

Ay = 3.14 sin (30) = 1.57 m

Vector B is resolved into its vertical and horizontal components,

Bx = 2.71 cos(60) = 1.355 m

By = ₋2.71 sin (60) = -2.35 m

C = A + B

  = (2.71+1.355) x + (1.57 -2.35) y

 = 4.064 i - 0.78 j

the magnitude of C = √(4.06)² + ( 0.78)² = 4.13 m

The angle, tan α = 0.78 / 4.06

 α = 10.8°

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select the set of quantum numbers that represents each electron in a ground‑state bebe atom.

Answers

The electron configuration of a ground-state Be atom is 1s² 2s². Each electron in the atom is described by a set of quantum numbers, including:

Principal quantum number (n): The first electron has n = 1 and the second electron has n = 2.

Azimuthal quantum number (l): For n = 1, the only possible value of l is 0 (s orbital), and for n = 2, the possible values of l are 0 (s orbital) and 1 (p orbital). Therefore, the first electron has l = 0 and the second electron has l = 0 or l = 1.

Magnetic quantum number (m): For l = 0, the only possible value of m is 0, and for l = 1, the possible values of m are -1, 0, and 1. Therefore, the first electron has m = 0, and the second electron has m = -1, 0, or 1.

Spin quantum number (s): Each electron has s = +1/2 or -1/2.

Thus, the set of quantum numbers for each electron in a ground-state Be atom is:

Electron 1: n = 1, l = 0, m = 0, s = +1/2 or -1/2

Electron 2: n = 2, l = 0 or 1, m = -1, 0, or 1, s = +1/2 or -1/2

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If arrivals occur according to the Poisson distribution every 20 minutes, then which is NOT true?
a.
λ = 20 arrivals per hour
b.
λ = 3 arrivals per hour
c.
λ = 1/20 arrivals per minute
d.
λ = 72 arrivals per day

Answers

If arrivals occur according to the Poisson distribution every 20 minutes, the statement that is NOT true is option, d: λ = 72 arrivals per day.

Given that arrivals occur according to the Poisson distribution every 20 minutes, we need to convert the rate to the appropriate time unit for each option:

a. λ = 20 arrivals per hour: This is true. The rate of 20 arrivals per hour is consistent with arrivals occurring every 20 minutes.

b. λ = 3 arrivals per hour: This is true. The rate of 3 arrivals per hour is consistent with arrivals occurring every 20 minutes.

c. λ = 1/20 arrivals per minute: This is true. The rate of 1/20 arrivals per minute is consistent with arrivals occurring every 20 minutes.

d. λ = 72 arrivals per day: This is NOT true. Since arrivals occur every 20 minutes, we need to convert the rate to arrivals per day. There are 24 hours in a day, and since arrivals occur every 20 minutes, there are 60 minutes / 20 minutes = 3 sets of 20 minutes in an hour. Therefore, there are 24 hours * 3 = 72 sets of 20 minutes in a day. The rate should be λ = 72 arrivals per day, not λ = 72 arrivals per day.

Therefore, option d is the statement that is NOT true.

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A 0.6 kg piece of metal displaces 1 liter of water when submerged. What is its density?

Answers

The density of the metal can be calculated using the formula:

density = mass / volume

where mass is the mass of the metal and volume is the volume of water displaced by the metal.

Given that the mass of the metal is 0.6 kg and it displaces 1 liter (1000 cubic centimeters) of water, we can substitute these values into the formula:

density = 0.6 kg / 1000 cubic centimeters

Simplifying, we get:

density = 0.0006 kg/cubic centimeter

Therefore, the density of the metal is 0.0006 kg/cubic centimeter.

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Which of the following states the relevance of the first law of thermodynamics to biology?
A) Energy is destroyed as glucose is broken down during cellular respiration.
B) Energy can be freely transformed among different forms as long as the total energy is conserved.
C) Photosynthetic organisms produce energy in sugars from sunlight.
D) The total energy taken in by an organism must be greater than the total energy stored or released by the organism.
E) Living organisms must increase the entropy of their surroundings.

Answers

The relevance of the first law of thermodynamics to biology. Energy can be freely transformed among different forms as long as the total energy is conserved. Correct option is B.

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transformed from one form to another. This principle is relevant to biology because living organisms are constantly converting energy from one form to another.

For example, photosynthetic organisms convert light energy into chemical energy stored in glucose molecules, which can then be used by other organisms through cellular respiration to produce ATP, the main source of energy for cellular processes.

The first law of thermodynamics also applies to other energy transformations that occur in biological systems, such as the breakdown of food molecules during digestion and the use of stored energy to power muscle contractions. Overall, this law emphasizes the importance of energy conservation and efficient energy use in biological systems.

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the big bang theory is supported by two major lines of evidence that alternative models have not successfully explained. what are they? the big bang theory is supported by two major lines of evidence that alternative models have not successfully explained. what are they? (1) the universe is expanding (2) the observed ratio of spiral to elliptical galaxies in the universe (1) the existence and specific characteristics of the observed cosmic microwave background (2) the observed overall chemical composition of the universe (1) the episode of inflation thought to have occurred in the early universe (2) the separation of gravity and the other forces at the end of the planck era (1) the early universe was hot and dense (2) we see distant galaxies as they were in the distant past

Answers

The two major lines of evidence that support the Big Bang theory and have not been successfully explained by alternative models are:

The existence and specific characteristics of the observed cosmic microwave background (CMB): The CMB is a form of radiation that fills the entire universe and is believed to be the leftover heat from the early universe when it was hot and dense. The Big Bang theory predicts the existence of the CMB and its specific characteristics, such as its uniformity and specific temperature fluctuations, which have been observed and confirmed through various experiments.

The observed overall chemical composition of the universe: The Big Bang theory predicts that the early universe was primarily composed of hydrogen and helium, with trace amounts of other elements. Observations of the chemical composition of the universe have confirmed this prediction and provide strong support for the Big Bang theory.

The other options provided in the question are not accurate as they do not represent the two major lines of evidence that support the Big Bang theory. The expansion of the universe is actually a consequence of the Big Bang theory, rather than evidence for it. The observed ratio of spiral to elliptical galaxies is not a major line of evidence for the Big Bang theory. The episode of inflation and separation of gravity from other forces are both aspects of the Big Bang theory itself and not independent lines of evidence supporting it. Lastly, the statement "the early universe was hot and dense" is a prediction of the Big Bang theory and not a line of evidence supporting it.

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