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Rom4ik [11]
2 years ago
13

A toy car (0.50 kg) runs on a frictionless track and has an initial kinetic energy of 2.2 J, as the drawing shows. The numbers b

eneath each hill give the heights of the hills. Over which of the hills will the car coast?
Physics
1 answer:
steposvetlana [31]2 years ago
6 0

Answer:

Height will be equal to 0.4489 m

Explanation:

We have given mass of the toy m = 0.50 kg

Initial kinetic energy K = 2.2 J

We have to fond the height of the hill over which car roast

When car will roast the hill its kinetic energy will be converted into potential energy and at maximum height all kinetic energy will be converted into potential energy

So at maximum height mgh=K

Acceleration due to gravity g=9.8m/sec^2

So 0.50×9.8×h = 2.2

h = 0.4489 m

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You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g
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Answer:

\displaystyle t=\frac{2v_o}{g}

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

\displaystyle y=\frac{gt^2}{2}

The horizontal distance is

x=v_ot

We need to find the time when both distances are equal, thus

\displaystyle \frac{gt^2}{2}=v_ot

Simplifying by t

\displaystyle \frac{gt}{2}=v_o

Solving for t

\displaystyle \boxed{t=\frac{2v_o}{g}}

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In the lab activity, you will examine sound waves as they are emitted from a moving source. Predict what will happen to the soun
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Which equation is most likely used to determine the acceleration from a velocity vs:time graph?
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2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

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q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

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Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

5 0
2 years ago
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