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docker41 [41]
3 years ago
10

As speed increases, what happens to potential energy ?

Physics
2 answers:
Charra [1.4K]3 years ago
8 0

Answer:

As the Speed increases , the potential energy decreases

dlinn [17]3 years ago
4 0

Answer:

Since the object only has so much total internal energy, and the faster an object moves the larger the kinetic energy is, hence the potential energy will decrease as a result of speed increasing.

Explanation:

Hope this helped!

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What is the speed of a 55.0 kg skydiver who has 7.81 x 104 J of kinetic energy?
KATRIN_1 [288]

Answer:

v=53.3m/s

Explanation:

Ek=1/2mv²

7.81×10⁴=1/2×55.0v²

v= the square root of 7.81×10⁴/0.5×55.0

v=53.3m/s

5 0
4 years ago
How are the 1st and 2nd Laws of Motion connected?
yuradex [85]

Answer:

Newton's First Law states that an object with no net force acting on it remains at rest or moves with constant velocity in a straight line. Newton's Second Law states that the acceleration of an object is directly proportional to the net force on the object and inversely proportional to the mass of the object.

Explanation:

hope it helps, pls mark as brainliest

4 0
4 years ago
If an object falls and ends with a velocity of 58.8 m/s, how long was the<br><br> object falling?
tensa zangetsu [6.8K]

Answer:

the time taken for the object to fall is 6 s.

Explanation:

Given;

final velocity of the object, v = 58.8 m/s

initial velocity of the object, u = 0

The height of fall of the object is calculated as;

v² = u² + 2gh

v² = 2gh

h = \frac{v^2}{2g} \\\\h = \frac{(58.8)^2}{2(9.8)} \\\\h = 176.4 \ m

The time to fall through the height is calculated as;

h =ut+  \frac{1}{2} gt^2\\\\h = 0 +  \frac{1}{2} gt^2\\\\h =  \frac{1}{2} gt^2\\\\t= \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 176.4}{9.8} } \\\\t = 6 \ s

Therefore, the time taken for the object to fall is 6 s.

6 0
3 years ago
A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the
Tema [17]

The cat fell 1.0 m from the ground.

Using the formula

h = v_{oy}t - \frac{gt^2}{2}

Here, v_{oy} = 0, \; h = 1.0 \; m.

Solving for t, the time it spent in the air is

t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s


5 0
3 years ago
Pls helppppppp I don’t understand this at all
Georgia [21]

Momentum is mass times velocity of the object

p = m • v

p = (7700 kg)(14 m/s)

p = 107809 kg m/s

3 0
3 years ago
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