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grandymaker [24]

2 years ago

13

A person is using a rope to lower a 8.0-N bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the forc

e exerted by the rope on the bucket

1 answer:

Stells [14]2 years ago

8 0

The magnitude of the force exerted by the rope on the bucket is 8N.

To find the answer, we need to know about the force.

<h3>What's force?</h3>

According to Newton's second law, force is expressed as rate of change of momentum.
It can be written as mass × acceleration.

<h3>What's the force experienced by an object, when it moves with constant velocity?</h3>

When an object moves with constant velocity, its acceleration is zero.
Since acceleration is zero, so the force experienced by it is zero.

Thus, we can conclude that there's no force on the bucket by the person as he is lowering it by constant velocity. But force on the bucket by the rope is 8N.

Learn more about the force here:

brainly.com/question/12785175

#SPJ4

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A tennis ball travels the length of the court 24m in 0.5 s find its average speed

melomori [17]

Speed = distance/time

4 0

3 years ago

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A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago

Which properties do metalloids share with metals?

Inga [223]

Metals are not brittle so it can’t be the first one or the third one, both metalloids and metals are shiny so it can’t be the second one. Therefore, it would be the last one because both metalloids and metals are shiny and both are solids at room temperature because it is not a high enough melting point.

ANSWER: Both are shiny and are solid at room temperature.

4 0

3 years ago

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What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f

Usimov [2.4K]

Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

$\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s$

Therefore, the rotational speed of the ring module have to be 0.92 rad/s

3 0

3 years ago

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0° with the horizo

Aleks [24]

Answer: Given mass of man = 85kg

Distance crate was pushed = 4m

Force exerted on crate = 500N

Angle inclined = 20°

Let Total force exerted Ft = (mg * sin(20)) + 500

Ft =( 85*9.81* sin20°) + 500

Ft = 758N

Work done Wd = Ft * distance

Wd = 785 * 4 = 3.14 * 10^3J

3 0

3 years ago