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padilas [110]
3 years ago
5

A race car drives one lap around a race track that is 500 meters in length.

Physics
1 answer:
DIA [1.3K]3 years ago
4 0

Answer:

a) 0

Explanation:

A) since he is going around a race track, his initial position is 0m, and his ending position is also 0m, so his displacement is 0. Yes, his distance is 500m, but his displacement is 0, and he did not end up going anywhere for his final position.

B) Distance is total amount traveled, which in this case is 500m, but displacement is 0 as he did not change his position.

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Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
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A 1kg mass is thrown to a height of 2cm. what is the potential energy​
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  • Mass=m=1kg
  • Height=h=2cm=0.02m
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\\ \tt\hookrightarrow P.E=mgh

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A divot is created _____.
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To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo
Setler [38]

Answer:

Being an elastic object, rubber ball will be an ideal choice as it will bounce off the bowling pit and will experience a large change in momentum in comparison with the beanbag which will either slow down or come to a halt upon hitting a bowling pit. That is why rubber ball will experience a greater impulse and the bowling pin will experience the negative impulse of the rubber ball.

For Rubber Ball

Upon elastic collision it will reverses the direction and move with velocity equal or less then original

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For Beanbag

value of impulse will large if velocity is zero.

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Explanation:

8 0
3 years ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
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Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

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KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

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3 years ago
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