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3 years ago

A car has a speed of 2 m/s and a mass of 1500 kg. What is the cars kinetic energy?

Physics

2 answers:

Kay [80]3 years ago

5 0

Answer:

Explanation:

E = mv² / 2

E = (1500kg)(2 m/s)² / 2

E = 6000 J / 2

E = 3000 J

Ghella [55]3 years ago

4 0

Answer:

Explanation: The answer is B because the defining equation for kinetic energy is E=mv^2 / 2

e = (1500kg)(2 m/s)^2 / 2

e = 6000 J / 2

e = 3000 J = 3000 Joules

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2 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

Answer:

Thunderbird is 995.157 meters behind the Mercedes

Explanation:

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$

initial velocity $u= 71 m/s$ distance

Distance covered in the slowing down phase = $250 m$

$v^2-u^2=2as$

$a= \frac {(v^2-u^2)}{2s}$

$a = \frac {(0^2-71^2)}{(2 \times 250)}=-10.082 \ m/s^2$

$v=u+at$

$t= \frac {(v-u)}{a}$

$= \frac {(0-71)}{(-10.082)}=7.042 s$

$t_1=7.042 s$

The car is in the pit stop for 5s $t_2=5 s$

After restart it accelerates for 350 m to reach the earlier velocity 71 m/s

$a= \frac {(v^2-u^2)}{(2\times s)} = \frac{(71^2-0^2)}{(2 \times 370)} =6.81 \ m/s^2$

$v=u+at$

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$t_3=10.425 s$

total time= $t_1 +t_2+t_3=7.042+5+10.425=22.467 s$

Distance covered by the Mercedes Benz during this time is given by $s=vt=71 \times 22.467= 1595.157 m$

Distance covered by the Thunderbird during this time= $250+350=600 m$

Difference between distance covered by the Mercedes and Thunderbird

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Thus the Mercedes is 995.157 m ahead of the Thunderbird.

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3 years ago

If the acceleration of motorboat is 4m/s^2 and the motorboat stsrtsfrol.Rest what is velocity after 6.0 s

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Answer:

24 m/s

Explanation:

Using v = u + at where u = initial velocity of the motorboat = 0 m/s (since the boat starts from rest), a = acceleration = 4 m/s², t = time = 6 s and v = velocity of the motorboat after 6.0 s.

Substituting the values of the variables into the equation, we have

v = u + at

= 0 m/s + 4 m/s² × 6.0 s

= 0 m/s + 24 m/s

= 24 m/s

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3 years ago

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