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iris [78.8K]
3 years ago
6

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

ccelerations. (a) a = -4.0 m/s^2 (b) a = -8.0 m/s^2
Physics
1 answer:
seropon [69]3 years ago
4 0

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

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Assoli18 [71]
Nuclear to electromagnetic
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3 years ago
A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

7 0
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MA_775_DIABLO [31]

Answer:

Yes both = and - g can be felt by a rider in a roller coaster.

Explanation:

It is crucial to understand how we feel gravity in this case.

We humans have no sensory organs to directly detect magnitude and direction like some birds and other creatures, but then how do we we feel gravity?

When we stand on our feet we feel our weight due to the normal reaction of floor on our feet trying to keep us stand and our weight trying to crush us down. In an elevator we feel difference in our weight (difference magnitudes of gravity) but actually we are feeling the differences in normal reactions under different accelerations of the elevator.

In the case of roller coaster you will feel +g as you  sit on a chair in it, but will feel -g when you are in upside down position as roller coaster move.

When you are seated you will feel the normal reaction of seat on you giving you the feeling +g and the support of the buckles to stay in the roller coaster when you are upside down will give you the -g feeling.

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8 0
3 years ago
Amina has a mass of 48.4kg, what is her weight.please show calculation
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G=m×g

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I hope it's true .d

good luck

6 0
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A positively charged sphere with a charge of 8Q is separated from a negatively charged sphere -2Q by a distance r. There is an a
djverab [1.8K]

Answer:

Explanation:

For the first case , the expression for electrostatic force can be given by the following .

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F = K 16 Q² / r²

When they touch , some charge is neutralized . Net charge remaining

= 8Q - 2 Q = 6 Q

Charge on each sphere = 6Q/2 = 3 Q .

Force between them

F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²

F₁ / F = 9 / 16

F₁ = 9 F / 16 .

4 0
2 years ago
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