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shepuryov [24]
3 years ago
6

6. A ball rolled 125 m from the top of a hill to the bottom of a hill. How long

Physics
1 answer:
leva [86]3 years ago
7 0
Time = distance / speed
T = 125/ 5
T = 25 meters per second
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Write your answer to the question below.
ElenaW [278]

Answer:

Hi... Potential energy is converted to kinetic energy and kinetic energy is converted to potential energy

8 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Leya [2.2K]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

6 0
3 years ago
Electromagnetic induction means that moving a magnet through a loop of wire creates an electric current.
MAVERICK [17]
False. An Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.
4 0
3 years ago
Read 2 more answers
In physics what does 7.56 × 5.746 equal ?
lapo4ka [179]

Answer:

43.43

Explanation:

5.746 x 7.56 = 43.43976

As the least number of desimal is two so our awnser should contain two digits after the decimal point.

Ans: 43.43.

7 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
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