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Elina [12.6K]
3 years ago
12

assume that the brakes in your car create a constant deceleration regardless of how fast you are going. if you double your drivi

ng speed, how does this affect (a) the time required to come to a stop, and (b) the distance needed to stop?
Physics
2 answers:
Monica [59]3 years ago
5 0
If that happens to your car and you double the speed then it will take the car longer to come to a complete stop
Novay_Z [31]3 years ago
5 0

Answer:

Part a)

Time becomes double to stop it

Part b)

Distance to stop the car will become 4 times

Explanation:

Part a)

To find the time to stop the car we know that final speed is zero

so we will have

v_f = v_i + at

here we have

0 = v - at

t = \frac{v}{a}

Now since the speed is double

so the time to stop will also becomes double

Part b)

To find the stopping distance we know that final speed of the object must be zero

so we will have

v_f^2 - v_i^2 = 2 ad

d = \frac{0 - v^2}{2(-a)}

d = \frac{v^2}{2a}

since the speed is double so the distance to stop the car will becomes 4 times

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What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s
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What travels faster than anything anything else in the universe?
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At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex
telo118 [61]

Answer:

346 * 10⁶ m

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The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let F_{e} be the force of gravity exerted by the earth

and let F_{m} be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d,  is given by:

F = \frac{Gm_{1} m_{2} }{d^{2} }

Mass of the earth, m_{e} = 5.97 * 10^{24} kg

Mass of the moon, m_{m} = 7.348 * 10^{22} kg

Mass of the satellite, m_{s} = ?

F_{e}  = \frac{G*5.97 * 10^{24} M }{d^{2} }...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

F_{m}  = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }............................(2)

Equating equations (1) and (2)

\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }

(5.97 * 10^{24})(14.78 * 10^{16}  -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d +  5.97 * 10^{24}d^{2} =  7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\

Factorising out 10^{24}

1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0

Solving for d in the quadratic equation  above:

d = 346 * 10⁶ m

4 0
3 years ago
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