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jarptica [38.1K]
3 years ago
7

a car moving at 30.0 m/s slows uniformly to a speed of 10.0 m/s in a time of 5.00 s determine the. acceleration of the car

Physics
1 answer:
netineya [11]3 years ago
7 0

Answer: -4m/s^{2}

Explanation:

Acceleration a is defined as the variation of Velocity \Delta V in time \Delta t :  

a=\frac{\Delta V}{\Delta t}

In this case \Delta V=10m/s - 30m/s  and \Delta t=5 s.

Therefore:

a=\frac{10m/s - 30m/s}{5 s}

a=\frac{-20m/s}{5 s}

a=-4 m/s^{2}  Note the negative sign indicates the car is slowing

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A physics student stands on a cliff overlooking a lake and decides to throw a tennis ball to her friends in the water below. She
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The horizontal component of the velocity = 19.5 cos 36.5 = 15.67m /s

The vertical component = 19.5 sin36.5 = 11.6 m /s

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s = ut + 1/2 gt²

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As the building collapses, the volume of air inside the building decreases, while the mass of the air stays the same. This means
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Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black
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Answer:

833.4801043*10^6N on ear that is closer to Black hole.

13.83803929*10^6N On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

                                                 F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

                     r = 120km-\frac{6}{100000} km=119.99994km=119999.94m

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

                                      F = 833.4801043*10^6N

       That is tension of ear.

(2) Force on ear that is further from black hole.

                              F =\frac{m_{1}*m_{2}  }{r^2} G

                    m_{1}= 5*1.989*10^30kg is mass of Black hole.

                     m_{2}= 0.030kg is mass of ear that is close to black hole.

                    G = 6.679*10^-11m^3*kg^-1*s^-2

         this time r is further away from black hole so it would be.

                       r = 120km+\frac{6}{100000} km = 120000.06m

Plugging this all in we get

                          F = 13.83803929N

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.

5 0
3 years ago
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