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solmaris [256]
3 years ago
13

(HELP ASAP) A large truck and a small two door car are both traveling 60 miles per hour down the highway, which has more kinetic

energy?
Physics
1 answer:
Harrizon [31]3 years ago
8 0
First we know that the large truck would weight more so it's heavier. Heavier objects have more kinetic energy than lighter ones. The large truck would have more kinetic energy than the car.
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Accelerating car is a _______
Pavel [41]

Acceleration is the rate of change of velocity as a function of time. For example a car traveling at 50 km/hr starts to accelerate, 10 seconds after, its speed changes to 100 km/hr then the acceleration of the car during the time can be calculated as below: initial speed = 50 km/hr.

5 0
2 years ago
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A roller coaster car rapidly picks up speed as it
TiliK225 [7]
Acceleration = (change of speed) / (time for the change)

Change in speed = (22 - 4) = 18 m/s.
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8 0
3 years ago
A sled with no initial velocity accelerates at a rate of 3.2 m/s2 down a hill. How long does it take the sled to go 15 m to the
grin007 [14]
S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
Find t
15 = 1/2(3.2)t^2
15 = 3.2t^2/2
30 = 3.2t^2
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3 0
3 years ago
Describe the relationship between motion<br> and a reference point.
steposvetlana [31]

Answer:

An object is in motion when its distance from another object is changing. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving.

Explanation:

4 0
3 years ago
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A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
Stels [109]

Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

\Delta U=-12 J-(-28 J)=+16 J

3 0
3 years ago
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