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luda_lava [24]
3 years ago
7

Two people, one of mass 85 kg and the other of mass 50 kg, sit in a rowboat of mass 90 kg. With the boat initially at rest, the

two people, who have been sitting at opposite ends of the boat, 3.5 m apart from each other, now exchange seats. How far does the boat move?
Physics
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

0.11m

Explanation:

let's assume the boat is of uniform construction

Ignoring friction losses

Also assume the origin is at the end of the boat originally with the heavier person

the center of mass of the whole system will not change relative to the water when the two swap ends

Originally, the center of mass is

85[0] + 90[3.5/2] + 50[3.5] / (85 + 90 + 50) = 1.14m from the origin

after the swap, the center of mass is

50[0] + 90[3.5/2] + 85[3.5] / (85 + 90+ 50) = 1.030m from the origin

The center of mass has shifted

1.14-1.030 = 0.11m

as no external force acted on the system, the center of mass relative to the water will not change. The boat will therefore shift towards the end where the heavier person originally sat

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A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten
Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a)  Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height,  hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0
3 years ago
. (a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110
7nadin3 [17]

Answer:

(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m

(b) thermal energy was generated by friction is 1.88 x 10^{5} J

(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N

Explanation:

given information:

m = 750 kg

initial velocity, v_{0} = 110 km/h = 110 x 1000/3600 = 30.6 m/s\frac{30.6^{2} }{2x9.8}

initial height, h_{0} = 22 m

slope, θ = 2.5°

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?

according to conservation-energy

EP = EK

mgh = \frac{1}{2} mv_{0} ^{2}

gh = \frac{1}{2} v_{0} ^{2}

h = \frac{v_{0} ^{2} }{2g}

  = 47.6 m

(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?

thermal energy = mgΔh

                         = mg (h - h_{0})

                         = 750 x 9.8 x (47.6 - 22)

                         = 188160 Joule

                         = 1.88 x 10^{5} J

(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?

f d  = mgΔh

f = mgΔh / d,

where h = d sin θ, d = h/sinθ

therefore

f = (mgΔh) / (h/sinθ)

 = 1.88 x 10^{5}/(22/sin 2.5°)

 = 373 N

8 0
3 years ago
A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler
natka813 [3]

Answer:

9.8 m/s^2, downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity (downward)

According to Newton's second law,

F=ma

where F is the net force on the object and a is its acceleration. Rearranging for a,

a=\frac{F}{m}

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

a=\frac{mg}{m}=g

Therefore, the ball has a constant acceleration of 9.8 m/s^2 downward for the entire motion.

5 0
3 years ago
(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is 3.85 108 m. Find
Goshia [24]

Answer:

It took 1.28 seconds to his voice to reach the Earth via radio waves.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies, the highest being that of gamma rays, followed by X-rays, ultraviolet rays and the visible region , and those of lower frequencies, which correspond to infrared, microwave and radio waves.

Light propagates as electromagnetic wave in vacuum with a speed of 3x10^{8}m/s. Therefore, radio waves will have in vacuum the same speed.

Then, to know the time that it took for its voice, the next equation can be used:

c = \frac{d}{t}  (1)

Where c is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

t = \frac{d}{c} (2)

t = \frac{3.85x10^{8} m}{3x10^{8}m/s}

t = 1.28s

Hence, it took 1.28 seconds to his voice to reach the Earth via radio waves.

7 0
3 years ago
Please help me , I also have to show work on paper
kramer

Answer:

Choose B

Explanation:

Hope Can I help you

7 0
3 years ago
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