Answer:
Therefore, the mass of the aluminum chunk is 15 g or 0.015 kg.
Explanation:
We have a container with 120 g of steel and 150 g of water at 25°C.
Two metal objects are dropped into the container:
A 244 g cube of copper initially at 90°C
An unknown mass of aluminum initially at 5.0°C
The surprise result is that the final water temperature is still 25°C.
Since the final water temperature is the same, the heat lost by the hot copper must equal the heat gained by the cold aluminum and the container.
The heat lost by the copper can be calculated as:
Qcopper = copper * copper * ΔTcopper
= 0.244 kg * 385 J/(kg•K) * (90°C - 25°C) = 29.5 kJ
The heat gained by the aluminum and container can be calculated as:
Qtotal = maluminum * cpaluminum * ΔTaluminum + mcontainer * cpcontainer * ΔTcontainer
Setting this equal to 29.5 kJ and plugging in the given values:
29.5 kJ = maluminum * 900 J/(kg•K) * (25°C - 5°C) + 0.120 kg * 450 J/(kg•K) * (25°C - 25°C)
Solving for aluminum :
aluminum = 0.015 kg
aluminum = 15 g
Answer:Certainly, let's solve the problem.
We can use the principle of conservation of energy to solve this problem. The heat lost by the hot object will be equal to the heat gained by the cold object and the water.
The heat lost by the copper cube can be calculated as:
Q1 = m1 * c1 * (T1 - Tfinal)
where
m1 = 244 g = 0.244 kg (mass of copper cube)
c1 = 0.385 J/g°C (specific heat of copper)
T1 = 90°C (initial temperature of copper cube)
Tfinal = 25°C (final temperature of the system)
Substituting the values, we get:
Q1 = 0.244 * 0.385 * (90 - 25) = 21.38 J
Similarly, the heat gained by the aluminum chunk can be calculated as:
Q2 = m2 * c2 * (Tfinal - T2)
where
m2 = ? (mass of aluminum chunk)
c2 = 0.902 J/g°C (specific heat of aluminum)
T2 = 5.0°C (initial temperature of aluminum chunk)
Substituting the values, we get:
Q2 = m2 * 0.902 * (25 - 5.0) = 18.144 m2 J
Now, since the water temperature doesn't change, we can assume that the heat gained by the water is equal to the heat lost by the hot objects. Therefore, we can write:
Q1 + Q2 = mwater * cwater * (Tfinal - Tinitial)
where
mwater = 150 g = 0.15 kg (mass of water)
cwater = 4.184 J/g°C (specific heat of water)
Tinitial = 25°C (initial temperature of water)
Substituting the values, we get:
21.38 J + 18.144 m2 J = 0.15 * 4.184 * (25 - 25)
Simplifying the equation, we get:
m2 = 0.266 kg
Therefore, the mass of the aluminum chunk is 0.266 kg or 266 grams.
Explanation:
Is there a difference between shapes when plotting Uniform acceleration towards (+)directtion,Uniform acceleration towards (-)direction, Uniform deceleration towards (+) direction and Uniform deceleration towards (-) direction in displacement time graph.Can you draw the shapes for each type ?
Explanation:
Yes, there are differences in the shapes of position-time graphs for uniform acceleration and uniform deceleration in different directions. Let's consider each case separately:[tex]\hrulefill[/tex]
(1) - Uniform acceleration towards the positive direction:
In this case, the object is moving in the positive direction with a constant acceleration. The displacement-time graph will typically be a curve that starts from an initial position and shows a steady increase in displacement over time. The shape of the graph will depend on the specific acceleration value.
(2) - Uniform acceleration towards the negative direction:
In this case, the object is moving in the negative direction with a constant acceleration. The displacement-time graph will also be a curve, but it will show a steady decrease in displacement over time.
(3) - Uniform deceleration towards the positive direction:
In this case, the object is initially moving in the positive direction but is slowing down with a constant deceleration. The displacement-time graph will be a curve that starts with a positive slope and gradually levels off.
(4) - Uniform deceleration towards the negative direction:
In this case, the object is initially moving in the negative direction but is slowing down with a constant deceleration. The displacement-time graph will be a curve that starts with a negative slope and gradually levels off.
At time t = 0, a vessel contains a mixture of 14 kg of water and an unknown mass of ice in equilibrium at 0°C. The temperature of the mixture is measured over a period of an hour, with the following results: During the first 45 min, the mixture remains at 0°C; from 45 min to 60 min, the temperature increases steadily from 0°C to 2.0°C. Neglecting the heat capacity of the vessel, determine the mass of ice that was initially placed in the vessel. Assume a constant power input to the container. Answer is in kg.
The mass of ice that was initially placed in the vessel of neglected heat capacity was found to be 1.135 kg.
Given that,
Mass of water = 14 kg
Mass of ice = m kg
P = Power of the burner = dQ/dt
Rate of the heat given by the burner is constant.
In the first 45 min,
dQ/dt = mL/t1 = m x (80 x 4.2 x 10³)/45 min. (1)
From 45 to 60 min,
dQ/dt = (m+14) x δ(Η₂Ο) x Δθ / t2 =(m+14) x (4.2 x 10³) x 2/ 15 min. (2)
From (1) and (2)
m x (80 x 4.2 x 10³)/45 min = (m+14) x (4.2 x 10³) x 2/ 15 min
80m/3 = (m+14) x 2
80m = 6m + 84
m = 1.135 kg
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Question 1
At one section of a long pipe the velocity of the fluid is 1.6 m/s. At another section of the pipe the diameter is three times greater.
What is the velocity of the fluid at this section?
O 0.533 m/s
○ 4.80 m/s
O Not enough information to tell
O 0.178 m/s
Question 2
Three thermometers are placed in a closed, insulated box and are allowed to reach thermal equilibrium. One is calibrated in
Fahrenheit degrees, one in Celsius degrees, and one in Kelvins. If the Celsius thermometer reads -40 °C the Fahrenheit
thermometer would read -40°F.
True
False
Answer:
Answer 1: The answer is O 0.178 m/s.
Answer 2: True: But in this specific case where the Celsius temperature is -40, the Fahrenheit temperature will also be -40.
So, in short, the answer is:
-40 Celsius is equal to -40 Fahrenheit
what impact does liberal arts have on ensuring continued innovation
Answer:
Liberal arts education is important for ensuring continued innovation. It helps students think creatively, solve problems, collaborate effectively, and consider ethical factors. By exploring various subjects and adapting to new situations, liberal arts education equips individuals with the skills needed to generate new ideas and drive progress in different fields.
On March 21_occurs where in the length of the day and night are equal
On March 21, an equinox occurs where the length of the day and night are equal. An equinox happens when the Earth's axis is not tilted towards or away from the sun. During an equinox, the sun's rays are equally distributed across the Earth's surface.
The word equinox is derived from the Latin words "aequus" and "nox," which means "equal night."
The equinox occurs twice a year, around March 21 and September 21. During an equinox, the duration of day and night is equal across the entire world.
It means that every place on the planet experiences almost the same amount of daylight and darkness.
On March 21, the vernal equinox, marks the beginning of spring in the Northern Hemisphere, while in the Southern Hemisphere, it marks the beginning of autumn.
The equinox has been considered a significant day in many cultures throughout history. In many cultures, the equinox is celebrated as a time of renewal, rebirth, and fertility. It's also been considered a day of balance, where darkness and light are equal, and the world is in harmony.
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A fish is swimming in the ocean at a depth with 14 atm of absolute pressure. If the fish swims up towards the surface, to a depth
that is now one third where it started, what is the gauge pressure at this depth?
4.67 atm
O 5.67 atm
3.67 atm
1.25 pts
O 4.33 atm
The gauge pressure at the new depth is determined as 3.67 atm.
What is the gauge pressure?The gauge pressure at the new depth is calculated by applying the following formula.
The absolute pressure at the new depth is;
abs P = (1/3) x 14 atm
abs P = 4.67 atm
The gauge pressure at the new depth is calculated as;
Gauge pressure = Absolute pressure at the new depth - Atmospheric pressure
G P = 4.67 atm - 1 atm
G P = 3.67 atm
Therefore, the gauge pressure at the new depth is 3.67 atm.
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The work of forensic engineers who investigate fires _____. identifies what started the fire and where it stated identifies what started the fire and where it stated does not consider explosions does not consider explosions is relatively simple is relatively simple is usually done by the firefighting team
The work of forensic engineers who investigate fires involves identifying what started the fire and where it originated, and it does not solely focus on explosions. This process is not relatively simple and is typically conducted by specialized forensic teams rather than firefighting personnel.
Forensic engineers play a crucial role in investigating fires to determine their cause and origin. Their primary objective is to gather evidence and analyze it in order to understand the circumstances surrounding the fire incident.
1. Scene Assessment: Forensic engineers begin by assessing the fire scene. They examine the area to gather initial information about the fire's intensity, pattern, and potential sources.
2. Evidence Collection: Next, the investigators collect physical evidence from the fire scene. This may involve gathering debris, examining burn patterns, and documenting any signs of accelerants or other substances that could have contributed to the fire.
3. Documentation: Forensic engineers meticulously document their findings through photographs, sketches, and written notes. This documentation serves as a crucial reference throughout the investigation process.
4. Laboratory Analysis: Collected evidence is then analyzed in specialized laboratories. Forensic experts employ various techniques such as chemical analysis, microscopy, and other scientific methods to determine the cause of the fire.
5. Report Preparation: Once the analysis is complete, forensic engineers prepare detailed reports outlining their findings. These reports serve as valuable resources for insurance companies, legal proceedings, and future fire prevention efforts.
It is important to note that the work of forensic engineers primarily focuses on identifying the cause and origin of the fire. While they consider various possibilities, including explosions, their investigation is not limited to solely investigating explosions.
Moreover, the process of investigating fires is complex and requires specialized knowledge and expertise. It is typically carried out by dedicated forensic teams rather than the firefighting personnel.
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¿Cuál es el trabajo neto en J que se necesita para acelerar un auto de 1500 kg de 55 m/s a 65 m/s?
What is the net work in J required to accelerate a 1500 kg car from 55 m/s to 65 m/s?
The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J
How do i determine the net work done?First, we shall obtain the initial kinetic energy. Details below:
Mass (m) = 1500 Kginitial velocity (u) = 55 m/sInitial kinetic energy (KE₁) =?KE₁ = ½mu²
= ½ × 1500 × 55²
= 41250 J
Next, we shall final kinetic energy. Details below:
Mass (m) = 1500 KgFinal velocity (v) = 65 m/sFinal kinetic energy (KE₂) =?KE₂ = ½mv²
= ½ × 1500 × 65²
= 3168750 J
Finally, we shall determine the net work done. Details below:
Initial kinetic energy (KE₁) = 41250 JFinal kinetic energy (KE₂) = 3168750 JNet work done (W) =?W = KE₂ - KE₁
= 3168750 - 41250
= 3127500 J
Thus, the net work done is 3127500 J
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The gravitational force exerted on a solid object is 5.05 N as measured when the object is suspended from a spring scale as in Figure a. When the suspended object is submerged in water, the scale reads 3.88 N (Figure b). Find the density of the object. Answer is in kg/m^3.
Answer:
Here are the steps to solve this problem:
Determine the weight of the object in air:
F = ma
Weight (W) = Force (F) x gravitational field strength (g)
W = 5.05 N x 9.8 m/s^2 = 49.5 N
Determine the weight of the object submerged in water:
W' = 3.88 N x 9.8 m/s^2 = 38.0 N
The difference in weight is due to buoyant force:
Fb = Wa - W
Fb = 38.0 N - 49.5 N = 11.5 N
The buoyant force is equal to the weight of the water displaced:
Fb = ρwVg (where ρw is the density of water and V is the volume)
11.5 N = (ρwV)(9.8 m/s^2)
Solve for the volume and density of the object:
V = Fb/( ρwg) = 11.5 N/(1000 kg/m^3)(9.8m/s^2)
= 1.17 x 10^-3 m3
Density of object = mass/volume
ρ = W/V = 49.5 N/(1.17 x 10^-3 m3)
= 4.22 x 104 kg/m3
= 4220 kg/m3
So the density of the object is 4220 kg/m3.
In summary, by using Newton's second law, Archimedes' principle and definitions of weight, buoyant force and density, we were able to determine the density of the object based on measurements in air and water.
Explanation:
1.5 A man with a mass of 90,6 kg walks to the back of a train at a velocity of 1m/s while the train moves at a constant velocity of 36 km/h in a easterly direction 1.5.1 The weight of the man 1.5.2 The velocity of the train in m/s 1.5.3 The resultant velocity of the man 1.5.4 The distance the train has travelled in ten (10) minutes (1) (1) (2) (2)
The weight of the man is approximately 888.6 Newtons.
The velocity of the train is approximately 10 m/s.
The resultant velocity of the man is 11 m/s.
The train has traveled a distance of 6000 meters in ten (10) minutes.
To solve this problem1.5.1 The weight of the man can be calculated using the formula:
Weight = mass * acceleration due to gravity
Given:
Mass of the man (m) = 90.6 kg
Acceleration due to gravity (g) ≈ 9.8 m/s²
Weight of the man = 90.6 kg * 9.8 m/s² ≈ 888.6 N
Therefore, the weight of the man is approximately 888.6 Newtons.
1.5.2 The velocity of the train in m/s can be converted from its given velocity in km/h. Since 1 km/h is equal to 1000/3600 m/s, we can calculate:
Velocity of the train (v) = 36 km/h * (1000/3600) m/s ≈ 10 m/s
Therefore, the velocity of the train is approximately 10 m/s.
1.5.3 By taking into account the relative motion of the man and the train, it is possible to determine the man's final velocity. The resultant velocity will equal the vector sum of the man's walking velocity and the train's velocity because the man is traveling toward the rear of the train, which is moving eastward.
Given:
Velocity of the man relative to the train (v_man) = 1 m/s (backwards)
Resultant velocity of the man (v_resultant) = v_man + v_train
v_resultant = 1 m/s + 10 m/s = 11 m/s
Therefore, the resultant velocity of the man is 11 m/s.
1.5.4 The distance the train has traveled in ten (10) minutes can be calculated using the formula:
Distance = velocity * time
Given:
Time (t) = 10 minutes = 10 * 60 seconds = 600 seconds
Velocity of the train (v) = 10 m/s
Distance = 10 m/s * 600 seconds = 6000 meters
Therefore, the train has traveled a distance of 6000 meters in ten (10) minutes.
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A 66 kg driver gets into an empty taptap to start the day's work. The springs compress 2.3×10−2 m
. What is the effective spring constant of the spring system in the taptap?
Enter the spring constant numerically in newtons per meter using two significant figures
Explanation:
You want N/m
N = 66 * 9.81
m = 2.3 x 10^-2 m
66* 9.81 / 2.3 x 10^-2 = 28150 = 28 000 N/m to two S D
What is indicated by the following displacement time graph ?
Answer:
The object in motion is deaccelerating in the negative direction.
Explanation:
Interpret the given position-time graph.
Observing the graph we can determine the slope is negative and is gradually flattening out. Thus, we can conclude the object in motion is deaccelerating in the negative direction.
A liquid ( = 1.65 g/cm^3) flows through a horizontal pipe of varying cross section as in the figure below. In the first section, the cross-sectional area is 10.0 cm^2, the flow speed is 293 cm/s, and the pressure is 1.20 * 10^5 Pa. In the second section, the cross-sectional area is 3.00 cm^2. Answer parts a-b.
Answer:
a) What is the flow speed in the second section? Assume the pressure remains constant.
Given:
Section 1 area = 10.0 cm^2
Section 1 flow speed = 293 cm/s
Section 2 area = 3.00 cm^2
Pressure = 1.20 x 105 Pa (constant)
Since pressure remains constant, by Bernoulli's equation:
P1/ρ + 1/2*v12 = P2/ρ+ 1/2*v22
Since P1 = P2 and ρ (density) is constant:
v22 = 2*(v12 - v22)
v22 = 2*(293^2 - v22)
Solving for v2:
v2 = √(2*293^2) = 846 cm/s
b) What is the kinetic energy loss per second between the two sections?
Kinetic energy = 1/2 * mass * velocity^2
Mass flow rate = density * area * velocity
= 1.65 g/cm^3 * 10.0 cm^2 * 293 cm/s = 485.5 g/s
Kinetic energy loss = 1/2 * (485.5 g/s) * (293^2 - 846^2) cm^2/s^2
= 1/2 * (485.5)*(86124 - 716256)
= 20617 J/s
So the kinetic energy loss per second between the two pipe sections is 20617 J/s.
Explanation:
Chapter 3
#1) State Kepler's three laws in your own words.
#2) Write out Newton's three laws of motion in terms of what happens with the momentum of objects.
3) According to Kepler's second law, where in a planet's orbit would it be moving the fastest?
Where would it be moving the slowest?
#4) The gas pedal, the brakes, and the steering wheel all have the ability to accelerate a car- how?
#5) Explain how a rocket can propel itself using Newton's third law.
#6) A certain material has a mass of 565 g while occupying 50 cm³3 of space. What is this material?
(Hint, Use Table 3.1)
#7) What was the great insight Newton had regarding Earth's gravity that allowed him to develop the
universal law of gravitation?
Kepler's three laws are:
Law 1: Planets move in oval-shaped paths around the Sun, and the Sun is at one of the special points inside this oval shape.Law 2: When a planet gets closer to the Sun, it moves faster. When it gets farther away, it moves slower.Law 3: When a planet takes longer to orbit the Sun, its average distance from the Sun increases.What is Newton's three laws of motionThe Newton's three laws of motion is one that how momentum works.
Law 1: If an object is not moving, it will stay still, and if it is moving, it will keep moving in the same way unless something outside of it makes it stop or change direction.
Law 2: When you push or pull on an object, the object's momentum changes. The faster you push or pull, the more the momentum changes.
Law 3: When you push or pull something, it pushes or pulls back with the same amount of force.
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A collapsible plastic bag (figure below) contains a glucose solution. If the average gauge pressure in the vein is 1.21 103 Pa, what must be the minimum height h of the bag in order to infuse glucose into the vein? Assume the specific gravity of the solution is 1.01. Answer is in h = _ m.
The minimum height (h) of the bag needed to infuse glucose into the vein is approximately 0.1235 meters.
How to solve for the minimum heightThe hydrostatic pressure is given by:
P = ρgh
Where:
P is the pressure,
ρ is the density of the fluid,
g is the acceleration due to gravity, and
h is the height of the fluid column.
Then we will have
h = P / (ρg)
h =[tex](1.21 * 10^3 Pa) / (1.01* 10^3 kg/m^3 * 9.8 m/s^2)[/tex]
h = 0.1235 meters
Therefore, the minimum height (h) of the bag needed to infuse glucose into the vein is approximately 0.1235 meters.
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A satellite weighing 5,400 kg is launched into orbit 3.6400 x 107 m above the center of the earth.
The mass of Earth is 6.0 × 1024 kg. The gravitational constant is 6.673 × 10–11 N•m2/kg2.
The gravitational force of Earth on the satellite is ___
Group of answer choices
9.1 x 10^4
1.6 x 10^3
2.1 x 10^6
Answer:
[tex]\tt F=1.63*10^3 N[/tex]
Explanation:
Gravitational force is defined as the force of attraction between two objects with mass. It is a fundamental force of nature, and it is what keeps us on the ground and what keeps the planets in orbit around the Sun.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers
For the Question:
We can use the following formula to calculate the gravitational force between the Earth and the satellite:
[tex]\boxed{\tt F =\frac{ G * M * m }{ r^2}}[/tex]
Where:
F is the gravitational force
G is the gravitational constant[tex]\tt (6.673 * 10^{-11} Nm^2/kg^2)[/tex]
M is the mass of the Earth [tex]\tt (6.0 * 10^24 kg)[/tex]
m is the mass of the satellite[tex]\tt (5,400 kg)[/tex]
r is the distance between the satellite and the center of the Earth [tex]\tt (3.6400 * 10^7 m)[/tex]
Plugging in these values, we get the following:
[tex]\tt F = \frac{6.673 * 10^{-11} * 6.0 * 10^{24}* 5,400 }{ (3.6400 * 10^7 )^2}[/tex]
[tex]\tt F=1.63*10^3 N[/tex]
Therefore, answer is [tex]\tt F=1.63*10^3 N[/tex]
A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 15.0°C. It is completely filled with turpentine at 15.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 85.0°C. (The average linear expansion coefficient for aluminum is 24 ✕ 10^−6°C^−1, and the average volume expansion coefficient for turpentine is 9.0 ✕ 10^−4°C−1.) Answer parts a-c.
Answer:
a) Calculate the change in the radius of the cylinder between 15.0°C and 85.0°C.
Given:
Depth of cylinder = 18.0 cm = 0.180 m
Average linear expansion coefficient for aluminum = 24 x 10^-6 /°C
Temperature change = 85 - 15 = 70 °C
Change in radius = (initial radius) x (linear expansion coefficient ) x (temperature change)
= (0.180/π) x (24 x 10^-6) x (70)
=2.16 x 10^-4 m = 0.0216 mm
b) Calculate the change in volume of the turpentine between 15.0°C and 85.0°C.
Given:
Initial volume of turpentine = 2.000 L
Average volume expansion coefficient for turpentine = 9.0 x 10^-4 /°C
Temperature change = 85 - 15 = 70 °C
Change in volume = (initial volume) x (volume expansion coefficient) x (temperature change)
= 2.000 L x (9.0 x 10^-4) x 70
= 0.126 L
c) Will any turpentine overflow? Explain your reasoning.
No turpentine will overflow because the increase in the radius of the cylinder is greater than the increase in the volume of the turpentine.
The cylinder radius increases by 0.0216 mm (part a) while the volume of turpentine increases by only 0.126 L (part b). This indicates the expanded cylinder can accommodate the increased volume of turpentine, so no overflow will occur.
Explanation:
if wrong im sorry
A cube of wood having an edge dimension of 20.9 cm and a density of 655 kg/m3 floats on water. Answer parts a-b.
Answer:
Mass of wood = .209^3 m^3 * 655 kg/m^3 = 5.98 kg
area of wood = ,209^2 = .04368 m^3
.04368 * h * 1000 = 5.98 where h is water displaced
h = .137 m = 13.7 cm
a) distance from wood to water = 20.9 - 13.7 = 7.2 cm
Mass of 7.2 cm of wood (wood out of water)
M = 655 kg /m^3 * .209^2 m^2 * .0072 m = .206 kg
.206 kg of Pb must be added to wood for submersion
Shannon and Chris push on blocks with identical force. SHannon's block is twice as massive as Chris'. After pushing for 5 seconds, who did more work?
What must be your car's average speed in order to travel 225 km in 3.35 h ?
Explanation:
Rate X Time = Distance
Distance / Time = Rate
225 km / 3.35 hr = 67.2 km/hr
If you feel pain during weight training, what should you do?
Answer:
Explanation:
Stop. Take a 5-10 minute breather, stretch out like you're doing your pre-workout (warmups). Drink some water, lay down for 3 minutes.
Answer: I hope this helps you
Explanation:
If you experience pain during weight training, it’s important to stop the exercise immediately and assess the source of the pain. If the pain is localized and mild, you can try taking a short break and then continuing with a lighter weight or modified exercise. However, if the pain is severe or spreading, you should stop the workout altogether and seek medical attention as soon as possible. Remember to always listen to your body and use proper form and technique to prevent injury.
The diagram below shows snapshots of an oscillator at different times . What is the frequency of the oscillation ?
In the diagram tha shows snapshots of an oscillator at different times, the frequency of the oscillation is 0.555 Hz.
How to calculate the periodThe period of the oscillation is the time taken for the for the object to return to its original position. (ie. Displacement = 0). From the above snapshot,
Period of oscillation = 1.80s.
From here, finding the frequency is simple as, Frequency = 1/Period. Hence,
Frequency = 1/1.80
= 0.555 Hz (3 sf).
The frequency of the oscillation is indeed 0.555 Hz. The frequency represents the number of oscillations or cycles per second. In this case, the object completes approximately 0.555 oscillations per second.
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Someone goes to lift a crate that is resting on the bottom of the pool filled with water (density of water is 1000 kg/m^3). While
still submerged, only 310 N is required to lift the crate. The crate is shaped like a cube with sides of 0.25 m. What is the density of
the cube? Numerical answer is assumed to be in units of kg/m^3
Answer:
the density of the cube is approximately 2016.07 kg/m^3.
Explanation:
The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Let's first calculate the weight of the crate:
mass of crate = density * volume = density * (side length)^3 = density * 0.25^3 = 0.015625 * density
weight of crate = mass of crate * gravity = 0.015625 * density * 9.81 = 0.1530875 * density
where gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.
Since the crate is submerged in water, the buoyant force acting on it is:
buoyant force = weight of water displaced = density of water * volume of water displaced * gravity
The volume of water displaced is equal to the volume of the cube, which is 0.25^3 = 0.015625 m^3. Therefore, the buoyant force is:
buoyant force = 1000 kg/m^3 * 0.015625 m^3 * 9.81 m/s^2 = 1.534453125 N
According to the problem, it takes 310 N to lift the crate while it is still submerged. This means that the net force acting on the crate is:
net force = lifting force - buoyant force = 310 N - 1.534453125 N = 308.465546875 N
This net force is equal to the weight of the crate:
net force = weight of crate = 0.1530875 * density
Therefore, we can solve for the density of the crate:
density = net force / 0.1530875 = 308.465546875 / 0.1530875 = 2016.06666667 kg/m^3
Rounding to the nearest hundredth, we get:
density ≈ 2016.07 kg/m^3
Therefore, the density of the cube is approximately 2016.07 kg/m^3
Hector stretches a spring with a spring constant of 3 N/m until it is extended by 50 cm. What is the elastic potential energy stored by the spring?
The elastic potential energy stored in the spring is 0.375 J.
The formula for elastic potential energy is:
E = 1/2 * k * x^2
where:
* E is the elastic potential energy in Joules
* k is the spring constant in N/m
* x is the distance the spring is stretched or compressed from its equilibrium position in meters
In this problem, we have:
* k = 3 N/m
* x = 0.5 m (50 cm)
Substituting these values into the formula, we get:
E = 1/2 * 3 * 0.5^2 = 0.375 J
Therefore, the elastic potential energy stored in the spring is 0.375 J.
Acceleration
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Study the motion map shown. Some of the vectors have been circled.
X
What do the circled vectors represent?
distance
speed
velocity
acceleration
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The circled vectors represents acceleration.
The last option is correct.
How do we explain?
We see in the first motion diagram the length of velocity vector is increasing this shows that the velocity is increasing in the magnitude with time so this is an accelerated motion in which a uniform acceleration must be in the same direction of velocity must be there.
We also notice in the second motion diagram the length of velocity vector is decreasing with time which shows the velocity is decreasing me magnitude which shows a constant deceleration and the direction of acceleration must be opposite to that of velocity.
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Select the correct answer.
1
If the distance between two objects is decreased to of the original
OA.
OB.
distance, how will it change the force of attraction between them?
O c.
The new force will be
The new force will be 100 times more than the original.
The new force will be 20 times more than the original.
1
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20
10
of the original.
If the distance between two objects is decreased to 1 10 of the original distance, the way it will change the force of attraction between them is (A) It will 100 times larger than the original force.
What is the force of attractionThe force of attraction between two objects is directly proportional to the inverse square of the distance between them and it is shown as:
F ∝ 1/r²
wher:
F is the force
r is the distance between the objects.
If the distance between the objects is decreased to 1/10 of the original distance (or 1/10r), one need to or can substitute this value into the equation. Hence it will be:
F' ∝ 1/(1/10r)²
So by Simplifying this expression, one have:
F' ∝ 1/(1/100r²)
F' ∝ 100r^2
From the above, one can see that the new force, F', is 100 times larger than the original force (F). Therefore, option A is the correct answer.
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See text below
If the distance between two objects is decreased to 1 10 of the original distance, how will it change the force of attraction between them? A. The new force will be 100 times more than the original. B. The new force will be 20 times more than the original. C. The new force will be 1 20 of the original. D. The new force will be 1 100 of the original. E. The new force will be 1 10 of the original.
a light string is wrapped around the rim of a small hoop if you hold the free end of the string in the hoop is released from rest it will unwind and the hoop descends, what force(s) is/are causing a torque on the hoop?
a-tension
b-weight
c-friction
d-normal force
e-more than one option is correct
Answer:
E: More than one option is correct
Explanation:
Tension, weight, and friction produce torques on the hoop, while the normal force does not.
A roller coaster, travelling with an initial speed of 15 meters per second, decelerates uniformly at -7.0 meters per second squared to a full stop. Approximately how far does the roller coaster travel during its deceleration
The distance traveled by the rollercoaster during the deceleration is 16.1 m.
What is the distance travelled by the roller coaster?The distance traveled by the rollercoaster during the deceleration is calculated by applying the following formula.
v² = u² + 2as
where;
v is the final velocity of the rollercoasteru is the initial velocity of the rollercoastera is the acceleration of the rollercoasters is the distance traveled by the rollercoasterWhen the rollercoaster stops, the final velocity, v = 0
0 = u² + 2as
-2as = u²
s = u² / -2a
s = (15 m/s) ² / (-2 x - 7 m/s²)
s = 225 / 14
s = 16.1 m
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The temperature of a aluminum bar rises by 10.0°C when it absorbs 4.73 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of aluminum from these data. Answer is in kJ/kg · °C.
Answer:
Certainly! We can use the formula:
q = mcΔT
where q is the amount of heat absorbed, m is the mass of the aluminum bar, c is the specific heat capacity of aluminum, and ΔT is the change in temperature.
Substituting the given values, we get:
4.73 kJ = (0.525 kg) x c x (10.0°C)
Solving for c, we get:
c = 0.901 kJ/kg · °C
Therefore, the specific heat of aluminum is 0.901 kJ/kg · °C.
Explanation:
HELPP PLEASEEE!! Now, for each time period, look at the graphs of x, vx, and ax. Briefly describe what is happening for each of these variables during your identified time periods. To view any of these graphs in detail, double-click the graph. You’ll be able to view it in the Data Tool window.
We can see here that to briefly describe what is happening for each variable during the identified time periods, you can follow these steps:
Identify the variablesDefine the time periodsGather data: Collect the necessary data or information for each variable during the identified time periods.Analyze the dataDescribe the variable's behaviorWhat is variable?A variable is a characteristic, property, or quantity that can vary or change in value. In the context of data analysis, research, or experimentation, variables are used to represent the different factors or elements that are being studied or observed.
Variables can take on different types and play various roles in a study or analysis.
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