Answer:
a
The speed is [tex]s = 5.857 m/s[/tex]
b
The distance is [tex]D = 22.4 \ m[/tex]
Explanation:
From the question we are told that
The speed of the banana is [tex]v = 16 \ m/s[/tex]
The distance from my location is [tex]d = 8.2 \ m[/tex]
The time taken is [tex]t = 1.4 \ s[/tex]
The speed of the ice cream is
[tex]s = \frac{d}{t}[/tex]
substituting values
[tex]s = \frac{8.4}{1.4}[/tex]
[tex]s = 5.857 m/s[/tex]
The distance of separation between i and Valdimir is the same as the distance covered by the banana
So
[tex]D = v * t[/tex]
substituting values
[tex]D = 16 * 1.4[/tex]
[tex]D = 22.4 \ m[/tex]
If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.
- 2x
- 1/4
- 1/2
- 4x
Answer:
The correct option is;
- 4x
Explanation:
From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source
The inverse square law can be presented as follows;
[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]
As the distance, r, increases, the surface it covers also increases by the power of 2
Therefore, where the distance increases from r to 2·r, we have;
When, I, remain constant
[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]
The surface increases to 4·S by the inverse square law
Therefore, the correct option is 4 × x.
Calculate the potential difference across a 25-Ohm. resistor if a 0.3-A current is flowing through it.
V
Answer:7.5V
Explanation:
Ohm's law, V=IR
so, V=0.3×25
V=7.5V
Answer:
7.5 V
Explanation:
A mass m at the end of a spring vibrates with a frequency of 0.72 Hz . When an additional 700 g mass is added to m, the frequency is 0.64 Hz . Part A What is the value of m? Express your answer using two significant figures.
Answer:
The value of m is 2635.294 grams.
Explanation:
Let suppose that mass-spring system has a simple harmonic motion, to this respect the formula for frequency is:
[tex]f = \frac{\omega}{2\pi}[/tex]
Where [tex]\omega[/tex] is the angular frequency, measured in radians per second.
For a mass-spring system under simple harmonic motion, the angular frequency is:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
Where:
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass, measured in kilograms.
The following equation is obtained after replacing angular frequency in frequency formula:
[tex]f = \frac{1}{2\pi}\cdot \sqrt{\frac{k}{m} }[/tex]
As this shows, frequency is inversely proportional to the square root of mass. Hence, the following relationship is deducted:
[tex]f_{1}\cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{2}}[/tex]
If [tex]m_{2} = m_{1} + 700\,g[/tex], [tex]f_{1} = 0.72\,hz[/tex] and [tex]f_{2} = 0.64\,hz[/tex], the resulting expression is simplified and then initial mass is found after clearing it:
[tex]f_{1} \cdot \sqrt{m_{1}} = f_{2} \cdot \sqrt{m_{1}+700\,g}[/tex]
[tex]f_{1}^{2} \cdot m_{1} = f_{2}^{2}\cdot (m_{1} + 700\,g)[/tex]
[tex]\left(\frac{f_{1}}{f_{2}} \right)^{2}\cdot m_{1} = m_{1} + 700\,g[/tex]
[tex]\left[\left(\frac{f_{1}}{f_{2}}\right)^{2} - 1\right]\cdot m_{1} = 700\,g[/tex]
[tex]m_{1} = \frac{700\,g}{\left(\frac{f_{1}}{f_{2}} \right)^{2}-1}[/tex]
[tex]m_{1} = \frac{700\,g}{\left(\frac{0.72\,hz}{0.64\,hz} \right)^{2}-1}[/tex]
[tex]m_{1} = 2635.294\,g[/tex]
The value of m is 2635.294 grams.
someone please help me with this thanks
"A trooper is moving due south along the freeway at a speed of 28 m/s. At time t = 0, a red car passes the trooper. The red car moves with constant velocity of 40 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.9 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car?"
Answer:
24.83 m
Explanation:
Applying the equation of motion;
d = vt + 0.5at^2 ......1
Where;
d = distance
v = velocity
t = time
a = acceleration
For the trooper;
v = 28 m/s
a = 2.9 m/s^2
Substituting into equation 1;
d1 = 28t + 0.5(2.9t^2)
d1 = 28t + 1.45t^2
For the red car;
v = 40 m/s
a = 0
Substituting into equation 1
d2 = 40t
The difference in distance is;
d = d2 - d1
d = 40t - (28t + 1.45t^2)
d = 12t - 1.45t^2
The maximum distance is at d(d)/dt = 0
differentiating d;
d' = 12 - 2.9t = 0
2.9t = 12
t = 12/2.9 = 4.137931034482
t = 4.138 s
Substituting t into function d;
d(max) = 12(4.138) - 1.45(4.138^2)
d(max) = 24.8275862 = 24.83 m
the maximum distance ahead of the trooper that is reached by the red car is 24.83 m
The Z0 boson, discovered in 1985, is the mediator of the weak nuclear force, and it typically decays very quickly. Its average rest energy is 91.19 GeV, but its short lifetime shows up as an intrinsic width of 2.5 GeV. what is the lifetime of this particle?
Answer:
The lifetime of the particle is [tex]\Delta t = 2.6*10^{-25} \ s[/tex]
Explanation:
From the question we are told that
The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]
The intrinsic width is [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]
The lifetime is mathematically represented as
[tex]\Delta t = \frac{h}{\Delta E}[/tex]
Where h is the Planck's constant with a value of [tex]1.055*10^{-34} \ J\cdot s[/tex]
substituting values
[tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]
[tex]\Delta t = 2.6*10^{-25} \ s[/tex]
Complete the first and second sentences, choosing the correct answer from the given ones.
1. The water temperature in the dish depends on the A / B / C / D.
A. average kinetic energy of water molecules
B. total kinetic energy of water molecules
C. water mass. D. potential energy of the container with water
2. The internal energy of the water in the vessel is E / F / G.
E. potential energy of the vessel with water
F. average kinetic energy of water molecules
G. sum of kinetic energy and potential water molecules
Answer:
Hope this helps :)
Explanation:
1. A
2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)
A student throws a set of keys vertically upward to his fraternity brother, who is in a window 3.60 m above. The brother's outstretched hand catches the keys 1.60 s later. (Take upward as the positive direction. Indicate the direction with the sign of your answer.)With what initial velocity were the keys thrown?
Answer:
[tex]v_{i}=10.10 m/s[/tex]
Explanation:
The equation of the position is:
[tex]y=y_{i}+v_{i}t-0.5gt^{2}[/tex]
Where:
v(i) is the initial velocity
The initial position y(i) will be zero and the final position y = 3.60 m.
So, we just need to solve this equation for v(i).
[tex]v_{i}=\frac{y+0.5gt^{2}}{t}[/tex]
[tex]v_{i}=\frac{3.6+0.5*9.81*1.6^{2}}{1.6}[/tex]
[tex]v_{i}=10.10 m/s[/tex]
Therefore, the initial velocity is 10.10 m/s upwards.
I hope it helps you!
An airplane takes off a runway at a constant speed of 49m/s at constant angle 30 to the horizontal
Complete Question
An airplane takes off a runway at a constant speed of 49 m/s at constant angle 30 to the horizontal.How high (in meters ) is the airplane above the ground 13 seconds after takeoff?
Answer:
The height is [tex]H = 318.5 \ m[/tex]
Explanation:
From the question we are told that
The speed at which the plane takes off is [tex]u = 49 \ m/s[/tex]
The angle at which it takes off is [tex]\theta = 30 ^o[/tex]
The time taken is [tex]t = 13 s[/tex]
The vertical distance traveled is mathematically represented as
[tex]H = u sin \theta t[/tex]
Substituting values
[tex]H = (49) * sin (30) *13[/tex]
[tex]H = 318.5 \ m[/tex]
A person is swimming 1.1 m beneath the surface of the water in a swimming pool. A child standing on the diving board drops a ball into the pool directly above the swimmer. The swimmer sees the ball dropped from a height of 4.2 m above the water. From what height was the ball actually dropped?
Answer:
The actual height is [tex]A =3.158 \ m[/tex]
Explanation:
From the question we are told that
The depth of the person is [tex]d = 1.1 \ m[/tex]
The apparent height is [tex]D = 4.2 \ m[/tex]
Generally
The refractive index of water is [tex]n_w = 1.33[/tex]
The refractive index of the air is [tex]n_a = 1[/tex]
The apparent depth is mathematically represented as
[tex]D = A [\frac{n_w}{n_a} ][/tex]
substituting values
[tex]4.2 = A [\frac{1.33}{1} ][/tex]
=> [tex]A = \frac{4.2 }{1.33}[/tex]
[tex]A =3.158 \ m[/tex]
The ball was dropped at the height of "3.158 m". To understand the calculation, check below.
Refractive IndexAccording to the question,
Water's refractive index, [tex]n_w[/tex] = 1.33
Air's refractive index, [tex]n_a[/tex] = 1
Apparent height, D = 4.2 m
Person's depth, d = 1.1 m
We know the relation,
→ D = A[[tex]\frac{n_w}{n_a}[/tex]]
By substituting the values, we get
4.2 = A[[tex]\frac{1.33}{1}[/tex]]
By applying cross-multiplication,
A = [tex]\frac{4.2}{1.33}[/tex]
= 3.158 m
Thus the approach above is correct.
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A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the skin and structure of the balloon have a mass of 990 kg ? Neglect the buoyant force on the cargo volume itself. Assume gases are at 0∘C and 1 atm pressure (rhoair = 1.29 kg/m3, rhohelium = 0.179 kg/m3).
Answer:
The mass of the cargo is [tex]M = 188.43 \ kg[/tex]
Explanation:
From the question we are told that
The radius of the spherical balloon is [tex]r = 7.40 \ m[/tex]
The mass of the balloon is [tex]m = 990\ kg[/tex]
The volume of the spherical balloon is mathematically represented as
[tex]V = \frac{4}{3} * \pi r^3[/tex]
substituting values
[tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]
[tex]V = 1697.6 \ m^3[/tex]
The total mass the balloon can lift is mathematically represented as
[tex]m = V (\rho_h - \rho_a)[/tex]
where [tex]\rho_h[/tex] is the density of helium with a value of
[tex]\rho_h = 0.179 \ kg /m^3[/tex]
and [tex]\rho_a[/tex] is the density of air with a value of
[tex]\rho_ a = 1.29 \ kg / m^3[/tex]
substituting values
[tex]m = 1697.6 ( 1.29 - 0.179)[/tex]
[tex]m = 1886.0 \ kg[/tex]
Now the mass of the cargo is mathematically evaluated as
[tex]M = 1886.0 - 1697.6[/tex]
[tex]M = 188.43 \ kg[/tex]
Two parallel plates having charges of equal magnitude but opposite sign are separated by 21.0 cm. Each plate has a surface charge density of 39.0 nC/m2. A proton is released from rest at the positive plate. (a) Determine the magnitude of the electric field between the plates from the charge density.
Answer:
E = 3.45*10^-19 N/C
Explanation:
a) The electric field between two parallel plates id given by the following formula:
[tex]E=\frac{\sigma}{\epsilon_o}[/tex] (1)
where:
σ: surface charge density of the plates = 39.0nC/m^2
εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2
You replace these values in the equation (1):
[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]
The electric field in between the parallel plates is 3.45*10^-19 N/C
Convert from standard form to scientific notation:
0.00000013
A)1.3 x 10-7
B)13 x 108
C)1.3 x 107
D)13 x 10-8
A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of the string is 0.62 m, and the tension in the string when the ball is at the top of the circle is 4.0 N. What is v
Answer:
Explanation:
Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force
T + mg = m v² / r
4 + .25 x 9.8 = .25 x v² / .62
6.45 = .25 v² / .62
v² = 16
v = 4 m /s .
You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 85 % of atmospheric pressure. What "holding force" does the suction cup generate in N
Answer:
Force = 60.08 N
Explanation:
Given that
Diameter d = 30 mm
Holding pressure = 85 % of Atmospherics pressure
Solution
As we know that here 1 atm = 10⁵ N/m²
and pressure is known as force per unit area
pressure = [tex]\frac{F}{A}[/tex] ................1
put here value and we will get
F = [tex]0.85\times 10^5\times \frac{\pi}{4}\times 0.03^2\ N[/tex]
solve it we get
Force = 60.08 N
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:
5727Co: 56.936296u
5726Fe: 56.935399u
Express your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
A negligible amount of this energy goes to the resulting 5726Fe atom as kinetic energy. About 90 percent of the time, after the electron-capture process, the 5726Fe nucleus emits two successive gamma-ray photons of energies 0.140MeV and 1.70 102MeV in decaying to its ground state. The electron-capture process itself emits a massless neutrino, which also carries off kinetic energy. What is the energy of the neutrino emitted in this case?
Express your answer in millions of electron volts.
Answer:
Explanation:
⁵⁷Co₂₇ + e⁻¹ = ²⁷Fe₂₆
mass defect = 56.936296 + .00055 - 56.935399
= .001447 u
equivalent energy
= 931.5 x .001447 MeV
= 1.3479 MeV .
= 1.35 MeV
energy of gamma ray photons = .14 + .017
= .157 MeV .
Rest of the energy goes to neutrino .
energy going to neutrino .
= 1.35 - .157
= 1.193 MeV.
The self-referencing effect refers to ________.
In a Venn diagram, the separate circles contain characteristics unique to each compared and the intersection contains characteristics that are common to both items being compared. This Venn diagram compares the inner and outer planets. What belongs in the center section?
a. -Revolve around the Sun
-Rotate on an axis
-Generally have rings
b. -Revolve around the Sun
-Rotate on axis
-Generally have moons
c. -Rotate around the Sun
-Revolve on an axis
-Generally have moons
d. -Rotate around the Sun
-Revolve on an axis
-Generally have rings
Answer:
B. revolve around the sun
rotate on an axis
generally have moons
Explanation:
edge 2021
In a circuit, a 100.-ohm resistor and a 200.-ohm resistor are connected in parallel to a 10.0-volt battery.
Calculate the equivalent resistance of the circuit. [Show all work, including the equation and substitution with units.]
Answer:
Explanation:
The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]
What is equivalent resistance?The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.
What is equivalent resistance in series?Resistors are in series whenever the current flows through the resistors sequentially. It is given by
[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]
What is equivalent resistance in parallel?Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.
The equivalent resistance is the total resistance measured in a parallel. It is given by
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]
Given:
Resistor, [tex]R_{1} = 100 ohm[/tex]
Resistor, [tex]R_{2} = 200 ohm[/tex]
Voltage, [tex]V = 10 Volt[/tex]
Since, resistors are connected in parallel, the equivalent resistor is given by,
[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]
[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]
[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]
Hence, the equivalent resistor is [tex]66.66 ohm[/tex].
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You are moving a desk that has a mass of 36 kg; its acceleration is 0.5 m / s 2. What is the force being applied
Answer:
18 N
Explanation:
Force can be found using the following formula.
f= m*a
where m is the mass and a is the acceleration.
We know the desk has a mass of 36 kilograms. We also know that its acceleration is 0.5 m/s^2.
m= 36 kg
a= 0.5 m/s^2
Substitute these values into the formula.
f= 36 kg * 0.5 m/s^2
Multiply 36 and 0.5
f=18 kg m/s^2
1 kg m/s^2 is equivalent to 1 Newton, or N.
f= 18 Newtons
The force being applied is 18 kg m/s^2, Newtons, or N
A ball thrown horizontally from the top of a building hits the ground in 0.600 s. If it had been thrown with twice the speed in the same direction, it would have hit the ground in:________.
a. 4.0 s.
b. 1.0 s.
c. 0.50 s.
d. 0.25 s.
e. 0.125 s.
Answer:
none of the answers is correct, the time is the same t₁ = t₂ = 0.600 s
Explanation:
This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)
let's find the time to hit the ground
y = y₀ + I go t - ½ g t²
0 = y₀ - ½ g t²
t = √ 2y₀ / g
with the data from the first launch
y₀i = ½ g t²
y₀ = ½ 9.8 0.6²
y₀ = 1,764 m
with this is the same height the time to descend in the second case is the same
t₂ = 0.600 s
this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis
Therefore, none of the answers is correct, the time is the same
t₁ = t₂ = 0.600 s
6. When a positive charge is released and moves along an electric field line, it moves to a position of A) lower potential and lower potential energy. B) lower potential and higher potential energy. C) higher potential and lower potential energy. D) higher potential and higher potential energy.
Answer:
Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.
The positive charge is released from a point such that it will move along an uniform electric field to the position of lower potential and lower potential energy. Therefore, option (A) is correct,
When a positive charge (say +Q) is released from a point (say A) and moves in an uniform electric field to reach the point (say B), then some work is done on the charge. This work done is given as,
[tex]W=+Q(V_{A}-V_{B})[/tex]
Here, [tex]V_{A}[/tex] and [tex]V_{B}[/tex] are the potential differences between the points A and B respectively..
This means the charge is moving from higher potential to lower potential. And since it is moving along the uniform electric field, therefore the electric potential energy of charged system is decreased.
Thus, we conclude that on releasing the positive charge from a point, it starts moving along the electric field towards the direction of lower electric potential and lower electric potential energy. Hence, option (A) is correct.
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Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2. What is the magnitude of the force exerted on the middle cube by the left cube in this case
Answer:
24 Newtons
Explanation:
The force exerted in the middle cube needs to be enough to move the middle cube and the right cube with an acceleration of 2 m/s2.
The mass of those two cubes combined is 6 + 6 = 12 kg
So, using the following equation, we can find the force:
Force = mass * acceleration
Force = 12 * 2
Force = 24 Newtons
When you "crack" a knuckle, you suddenly widen the knuckle cavity, allowing more volume for the synovial fluid inside it and causing a gas bubble suddenly to appear in the fluid. The sudden production of the bubble, called "cavitation", produces a sound pulse---the cracking sound. Assume that the sound is transmitted uniformly in all directions and that it fully passes from the knuckle interior to the outside, at a distance of 0.29 m from your ear. If the pulse has a sound level of 61 dB at your ear, what is the rate at which energy is produced by the cavitation
Answer:If a wave y(x, t) = (6.0 mm) sin(kx + (600 rad/s)t + Φ) travels along a string, how much time does any given point on the string take to move between displacements y = +2.0 mm and y = -2.0 mm?
Explanation:
Two parallel, vertical, plane mirrors, 38.8 cm apart, face each other. A light source at point P is 30.1 cm from the mirror on the left and 8.7 cm from the mirror on the right.
(a) How many images of point P are formed by the mirrors?
(b) Find the distance from the mirror on the right to the two nearest images behind the mirror.
first nearest image=
second nearest image=
(c) Find the number of reflections of light rays for each of these images.
first nearest image=
second nearest image=
Answer:
Explanation shown below.
Explanation:
1.The number of images formed by 2 parallel mirrors is an infinite number of images.
2. The characteristics of a plane mirror is such that the object distance equals the image distance.
Hence the object distance is 8.7cm from the right; the image formed would be 8.7cm behind the mirror.
Now a second image is going to be formed by the left mirror which is going to have an image distance of 30.1cm behind the mirror.
Now this image would be reflected on the right side to form a new image which is going to be seen as 38.8 +30.1 = 68.9cm behind the right Mirror .
Hence the shortest distances are 8.7cm and 68.9cm
3. The number of reflections is infinite for both cases.
The red giant Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that large, we would be inside it!) Assume that it radiates like an ideal blackbody.a) If Betelgeuse were to radiate all of its energy at the peak-intensity wavelength, how many photons per second would it radiate?b) Find the ratio of the power radiated by Betelgeuse to the power radiated by our sun (at 5800 K).
Answer:
Explanation:
a )
Radius of the sun = .69645 x 10⁹ m .
600 times = 600 x .69645 x 10⁹ m
= 4.1787 x 10¹¹ m .
surface area A = 4π (4.1787 x 10¹¹)²
= 219.317 x 10²²
energy radiated E = σ A Τ⁴
= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴
= 100695 x 10²⁶ J
To know the wavelength of photon emitted
[tex]\lambda_mT= b[/tex]
[tex]\lambda_m= \frac{b}{T}[/tex]
= 2.89777 x 10⁻³ / 3000
= 966 nm
= 1275 /966 eV
1.32 x 1.6 x 10⁻¹⁹ J
= 2.112 x 10⁻¹⁹ J
No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹
= 47677.5 x 10⁴⁵
= .476 x 10⁵⁰ .
b )
energy radiated by our sun per second
E₂ = σ A 5800⁴
energy radiated by Betelgeuse per second
E₁ = σ x 600²A x 3000⁴
E₁ / E₂ = σ x 600²A x 3000⁴ / σ A 5800⁴
= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸
= 25.76 x 10⁸ x 10⁻⁵
= 25760 times .
A train starts from rest and accelerates uniformly, until it has traveled 5.6 km and acquired a velocity of 42 m/s. The train then moves at a constant velocity of 42 m/s for 420 s. The train then slows down uniformly at 0.065 m/s^2, until it is brought to a halt. What is the acceleration during the first 5.6 km of travel?
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
[tex]v_f^2 = v_i^2 + 2*a*( s_f - s_o )[/tex]
- We will plug in the given parameters in the equation of motion given above:
[tex]42^2 = 0^2 + 2*a* ( 5600 - 0 )\\\\1764 = 11,200*a\\\\a = \frac{1,764}{11,200} \\\\a = 0.1575 \frac{m}{s^2}[/tex]
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2
A block is supported on a compressed spring, which projects the block straight up in the air at velocity VVoj The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block?
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.
Answer:
the correct answer is B
Explanation:
We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.
For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.
Let's write the kinematic equation for the two bodies
The block. At the highest point of the path
y = - ½ g t2
The ball, in its vertical movement
y = vo t - ½ g t2
therefore the correct answer is B
A swimmer heading directly through a 200m wide river reaches the opposite shore in 6 min 40s. She is washed downstream 480 m. How fast can you swim in calm water?
Answer :v=480m400s=1.2ms
2002+4802=H2
The hypotenuse H=520m
A quicker way to get the length of the hypotenuse is to recognize that this is a simple 5–12–13 triangle where the sides are multiples of 5, 12, and 13:
5(40) = 200m, 12(40)= 480m, 13(40)= 520m
We know that the swimmer travelled 520 m in 400 seconds, so her average speed was:
VR=520m400sec= 1.3ms
hope i got it right!! xx
Explanation:
Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an upstairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?
Answer:
The velocity is [tex]v_2= 0.45 \ m/s[/tex]
Explanation:
From the question we are told that
The initial speed of the hot water is [tex]v_1 = 0.85 \ m/s[/tex]
The pressure from the heater [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]
The height of the hot water before flowing is [tex]h_1 = 0 \ m[/tex]
The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]
The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]
The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]
Apply Bernoulli equation
[tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]
Substituting values
[tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]
=> [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]
=> [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]
=> [tex]v_2= 0.45 \ m/s[/tex]
