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Anvisha [2.4K]
3 years ago
8

Determine the kinetic energy of a 1000kg rollercoaster car that is moving with a speed of 20.0 m/s

Physics
1 answer:
stich3 [128]3 years ago
7 0
KE = 1/2mv^2
KE = 1/2(1000)(20^2)
KE = 2.0 * 10^5 
or 
KE = 200,000

hope this helps :)
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What types of changes in motion cause acceleration?
xxTIMURxx [149]

Answer:

Change in velocity and direction over a specific period of time.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate the acceleration of an object.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is final and initial velocity respectively, measured in ms^{-1}

t is time measured in seconds.

Hence, the types of changes in motion that cause acceleration is a change in velocity and direction over a specific period of time.

6 0
3 years ago
What do wind and moving water have in common? Select one: They both have electrical energy They both have elastic potential ener
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They both have kinetic energy because they are both moving and have force.
Hope this helped! :)

7 0
3 years ago
Which clouds are often associated with thunder and lightning?
ehidna [41]
Your answer is cumulonimbus clouds
3 0
2 years ago
6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t
Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

y = u sin \theta - \dfrac{1}{2}gt^2................(2)

from equation(1) and (2)

y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}..........{3}

12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}

\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29

\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2

u = \sqrt{10959.34}

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

8 0
3 years ago
A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the
asambeis [7]
EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
7 0
2 years ago
Read 2 more answers
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