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Tom [10]
3 years ago
10

Where does the basketball have the greatest gravitational potential energy?

Physics
1 answer:
Olenka [21]3 years ago
6 0
The basketball has the greatest gravitational potential energy at its, highest which is at the very top.
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A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial sp
Dimas [21]

By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s

Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:

mass m =  0.170 kg

initial speed u = 6 m/s.

Distance covered s = 61 m

To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V

To do this, let us first calculate the kinetic energy at which the ball move.

K.E = 1/2mU^{2}

K.E = 1/2 x 0.17 x 6^{2}

K.E = 3.06 J

The work done on the ball is equal to the kinetic energy. That is,

W = K.E

But work done = Force x distance

F x S = K.E

F x 61 = 3.06

F = 3.06/61

F = 0.05 N

From here, we can calculate the acceleration of the ball from Newton second law

F = ma

0.05 = 0.17a

a = 0.05/0.17

a = 0.3 m/s^{2}

To calculate the final velocity, let us use third equation of motion.

V^{2} = U^{2} + 2as

V^{2}  = 6^{2} + 2 x 0.3 x 61

V^{2} = 36 + 36

V^{2} = 72

V = \sqrt{72}

V = 8.485 m/s

Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.

Learn more about dynamics here: brainly.com/question/402617

5 0
2 years ago
A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res
VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

KE = PE

\frac{1}{2} mv^2 = mgh

v = \sqrt{2gh}

v = \sqrt{2(9.8)(2(55.8*10^{-2}))}

v = 4.67m/s

Therefore the minimum speed at the bottom is required to make the ball go over the top of the circle is 4.67m/s

4 0
3 years ago
Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo
telo118 [61]

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

6 0
3 years ago
You toss a coin into a wishing well full of a liquid denser than the coin. Which of the following could be true?
maks197457 [2]
If one of the answers are the coin will float that one would be correct.
8 0
4 years ago
Read 2 more answers
Draw Aufbua diagram of sodium ion​
Mazyrski [523]

Answer:

Aufbau diagram of sodium ion:

2s²,2p⁶,3s¹

5 0
3 years ago
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