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ivann1987 [24]
3 years ago
7

Which of these is exhibiting kinetic energy? Which of these is exhibiting kinetic energy? the high-energy phosphate bonds of a m

olecule of ATP a space station orbiting Earth a rock on a mountain ledge a person sitting on a couch while watching TV an archer with a flexed bow Submit
Physics
1 answer:
Wittaler [7]3 years ago
7 0

Answer:

A space station orbiting

Explanation:

As we know that kinetic energy associated with the motion of particle and it is given as

m= Mass

V =Velocity

KE=\dfrac{1}{2}mV^2

If V= 0 ,means velocity of particle is zero then it kinetic energy will be zero.

A)

At ATP molecule does not have any movement that is why it kinetic energy is zero.

B)

A space station have velocity that is why it have kinetic energy.

C)

A rock on mountain ledge do not have kinetic energy .Because it is not moving.

D)

When person sitting it means that it dies not have any motion that is why it does not have kinetic energy.

You might be interested in
Which of the following best describes the function of a plant's flowers?
noname [10]

Answer: The function of a plant's flower is reproduce. Since the flowers are the reproductive organs of plant, they mediate the joining of the sperm, contained within pollen, to the ovules contained in the ovary.

Explanation:

8 0
3 years ago
A 4.80 −kg ball is dropped from a height of 15.0 m above one end of a uniform bar that pivots at its center. The bar has mass 7.
Margarita [4]

Answer:

h = 13.3 m

Explanation:

Given:-

- The mass of ball, mb = 4.80 kg

- The mass of bar, ml = 7.0 kg

- The height from which ball dropped, H = 15.0 m

- The length of bar, L = 6.0 m

- The mass at other end of bar, mo = 5.10 kg

Find:-

The dropped ball sticks to the bar after the collision.How high will the other ball go after the collision?

Solution:-

- Consider the three masses ( 2 balls and bar ) as a system. There are no extra unbalanced forces acting on this system. We can isolate the system and apply the principle of conservation of angular momentum. The axis at the center of the bar:

- The angular momentum for ball dropped before collision ( M1 ):

                                 M1 = mb*vb*(L/2)

Where, vb is the speed of the ball on impact:

- The speed of the ball at the point of collision can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mb*g*H = 0.5*mb*vb^2

                                  vb = √2*g*H

                                  vb = ( 2*9.81*15 ) ^0.5

                                  vb = 17.15517 m/s

- The angular momentum of system before collision is:

                                  M1 = ( 4.80 ) * ( 17.15517 ) * ( 6/2)

                                  M1 = 247.034448 kgm^2 /s

- After collision, the momentum is transferred to the other ball. The momentum after collision is:

                                  M2 = mo*vo*(L/2)

- From principle of conservation of angular momentum the initial and final angular momentum remains the same.

                                 M1 = M2

                                 vo = 247.03448 / (5.10*3)

                                 vo = 16.14604 m/s

- The speed of the other ball after collision is (vo), the maximum height can be determined by using the principle of conservation of energy:

                                  ΔP.E = ΔK.E

                                  mo*g*h = 0.5*mo*vo^2

                                  h = vo^2 / 2*g

                                  h = 16.14604^2 / 2*(9.81)

                                  h = 13.3 m

3 0
3 years ago
A certain wave has a compressions and rarefactions.How should this wave be classified?
AlexFokin [52]
A) as a longitudinal wave
7 0
3 years ago
A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu
Pepsi [2]

Answer:

v = 10.89\ m/s

Explanation:

given,                          

radius of loop = 12.1 m                              

to find the minimum speed transverse by the rider to not to fall out upside down                                                                

centripetal force = \dfrac{mv^2}{r}

gravitational force  = m g

computing both the equation]

mg = \dfrac{mv^2}{r}

v = \sqrt{rg}

v = \sqrt{12.1 \times 9.8}

v = \sqrt{118.58}

v = 10.89\ m/s

5 0
3 years ago
A projectile is fired with an initial speed of 60.3 m/s at an angle of 34.2 above the horizontal on a long flat firing range.
amid [387]

Answer:

A.) H = 58.6 m

B.) T = 6.92 s

C.) 345.12 m

D.) V = 22.13 m/s

E.) Ø = 32.1 degree

Explanation:

Given that the

initial speed U = 60.3 m/s

Angle Ø = 34.2 degree

A.) At maximum height, final velocity V is equal to zero.

Using the third equation of motion under gravity.

V^2 = U sin Ø^2 - 2gH

Substitute for U and g. Where g = 9.8 m/s^2

0 = (60.3 sin 34.2)^2 - 2 × 9.8 × H

1148.78 = 19.6 H

H = 1148.78/19.6

H = 58.6 m

B.) To Determine the total time in the air, let us use the formula

V = UsinØ - gt

At maximum height, V = 0

t = UsinØ/g

Total time T = 2t

Therefore, T = 2UsinØ/g

T = (2 × 60.3 × sin 34.2)/9.8

T = 67.79/9.8

T = 6.92 s

C.) To determine the total horizontal distance covered which is the range, we will use second equation of motion.

S = UcosØT - 1/2gt^2

Where S = range R

g = 0, since the range is not a vertical distance

T = total time

Substitute all the parameters into the formula

R = 60.3 cos 34.2 × 6.92

R = 345.12 m

D.) After 1.2 s firing,

V = UsinØ - gt

Where t = 1.2 s

Substitute into the formula

V = 60.3 × sin34.2 - 9.8 × 1.2

V = 33.89 - 11.76

V = 22.13 m/s

Therefore the speed of the projectile 1.20 s after firing is 22.13 m/s

E.) The direction will be determined by using the formula

t = VsinØ/ g

Cross multiply

VsinØ = gt

Make SinØ the subject of formula

SinØ = gt/V

SinØ = (9.8×1.2)/22.13

Sin Ø = 11.76/22.13

Sin Ø = 0.53

Ø = sin^-1( 0.53 )

Ø = 32.1 degree

3 0
3 years ago
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