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3 years ago

WILL GIVE THAT CROWN THINGY!

Physics

2 answers:

frutty [35]3 years ago

3 0

Conservation of energy explains that energy can only be transferred between different forms of energy

yuradex [85]3 years ago

3 0

Conservation of energy explains that energy can only transfer

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Experiments and investigations must be ____. a. approved b. repeatable c. not reproducible d. accepted

deff fn [24]

Experiments and investigations must be B. Repeatable.

7 0

3 years ago

Read 2 more answers

The final velocity of car is 20 m/s. The car is accelerating at a rate of 2.5m/s over a 10 seconds period of time. What is the i

balu736 [363]

Answer:

-5m/s

Explanation:

Since

acceleration=final velocity-initial velocity/time

2.5m/s^2=20m/s- initial velocity/10s

2.5m/s^2×10s= 20m/s -initial velocity

25m/s=20m/s - initial velocity

Initial velocity=20m/s-25m/s

= -5m/s

4 0

3 years ago

A 92.0-kg skydiver falls straight downward with an open parachute through a vertical height of 325 m. The skydiver's velocity re

natka813 [3]

Answer: Workdone293.02KJ

Explanation: The equation to use to calculate Workdone = Change in KE + Change in PE

Assuming velocity is constant,KE becomes 0

Workdone= Change in PE=mg

W=92×9.8×325=293.02KJ

5 0

4 years ago

A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000

allochka39001 [22]

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

Mass of the car, m = 1700 kg
Velocity of the car, v = 21 m/s
Time of motion, t = 10 s
Additional energy provided by the engine, E₁ = 22,000 J
Energy used in overcoming friction, E₂ = 3,666.67 J

The change in the energy applied to the car is calculated as;

$\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta E = 18,333.33 \ J$

The final speed of the car is calculated as follows;

$\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2 = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s$

Thus, the final speed of the car at the given conditions is 30.1 m/s.

Learn more about change in kinetic energy here: brainly.com/question/6480366

4 0

3 years ago

A car is sitting motionless on top of a bridge. The bridge pushes up on the car with a force of 782 Newtons.

Evgesh-ka [11]

Answer: The car would move to the right with 150 n of force. the 150 n of force from friction and air resistance slows the car from moving to the right with 300 n of force.

Explanation:

3 0

3 years ago