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suter [353]
2 years ago
11

If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet

ic enegy
Physics
1 answer:
dezoksy [38]2 years ago
6 0
The kinetic energy of an object of mass m and velocity v is given by
K= \frac{1}{2} mv^2

Let's call v_i the initial speed of the car, so that its initial kinetic energy is
K_i =  \frac{1}{2} mv_i^2
where m is the mass of the car. 

The problem says that the car speeds up until its velocity is twice the original one, so 
v_f = 2 v_i
and by using the new velocity we can calculate the final kinetic energy of the car
K_f =  \frac{1}{2} mv_f^2 =  \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.
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Explanation:

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A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

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3 years ago
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