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Softa [21]
2 years ago
14

De donde eres responde y te doy corona

Physics
1 answer:
irakobra [83]2 years ago
8 0

soy de texas, united states

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How much potintial energy is in a closed system where the height of the object is 2 m, and the mass is 10 kg?
Lorico [155]

Answer:

D. 196 J

Explanation:

PE=mgh=10×9.8×2=196J

6 0
2 years ago
Read 2 more answers
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
___________________is the rate that an object moves in a certain direction.
AleksandrR [38]

Answer:

Velocity

Explanation:

<u>Velocity</u> is the rate that an object moves in certain direction.

3 0
1 year ago
For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro
Mariulka [41]

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

v = u + gt

Where v is the final velocity

u is the initial velocity

g is acceleration due to gravity (take g = 9.8 m/s²)

and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

Also, from one of the equations of kinematic for free fall motions,

h = ut + 1/2(gt²)

u = 0 m/s

From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore

h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

5 0
3 years ago
What happens to the mass and volume if density increases
nikitadnepr [17]
They all stay the same regardless
7 0
3 years ago
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