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Makovka662 [10]
3 years ago
6

An elevator of massmis initially at rest onthe first floor of a building. It moves upward,and passes the second and third floors

witha constant velocity, and finally stops at thefourth floor.The distance between adjacentfloors is h.What is the net work done on the elevatorduring the entire trip, from the first floor tothe fourth floor?
1. None of these.
2. W=-3m g h
3. W=-4m g h
4. W= 3m g h
5. W= 0correct
6. W= 4m g h
Physics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

To solve this problem it is necessary to apply the concepts of Work. Work is understood as the force applied to travel a determined distance, in this case the height. The force in turn can be expressed by Newton's second law as the ratio between mass and gravity, as well

W = mgh

Where,

m = mass

h = height

g = Gravitational constant

When it ascends to the second floor it has traveled the energy necessary to climb a height, under this logic, until the 4 floor has traveled 3 times the height h of each of the floors therefore

h = 3h

Replacing in our equation we have to

W = mgh\\W = mg(3h)\\W = 3mgh

The correct answer is 4.

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"Fig is attacted with answer"

Answer:

a) d = 33.72 m

b) v_{i} = 26 m/s

c) β = 71.08°

Explanation:

a)

When an object is thrown into the air under the effect of the gravitational force, the movement of the projectile is observed. Then it can be considered as two separate motions, horizontal motion and vertical motion. Both motions are different, so that they can be handled independently.

Given data:

time = t = 4.00 s

Height = h = 20 m

Angle = θ = 60°

Horizontal distance = d = ?

Using 2nd  equation of motion

h = v_{y_{f}}t + \frac{1}{2}gt^{2}

-20 = v_{y_{f}} (4) + 0.5(-9.8)(4)²

v_{y_{f}} (4) = 58.4

v_{y_{f}}  = 14.6 m/s

This is vertical component of velocity when the ball is on the roof. To calculate the Final velocity and horizontal component, we use

v_{f} = v_{y_{f}} / sinθ

v_{f} = 14.6 / sin 60

v_{f} = 16.86 m/s

v_{x_{f}} = v_{f}cosθ

v_{x_{f}} = 16.86 cos 60

v_{x_{f}} = 8.43 m/s

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d = v_{y_{f}} t

d = (8.43)(4)

d = 33.72 m

b)

We know the values of Landing angle, height of roof, time of flight. In part a, We calculate the landing velocity of the ball and also its horizontal and vertical component. As the ball followed the projectile path, and we know that in projectile motion the horizontal component of the velocity remain constant throughout his motion. So there is no acceleration along horizontal path.

So,

v_{x_{f}} = v_{x_{i}}

but the vertical component of velocity vary with and there is an acceleration along vertical direction which is equal to gravitation acceleration g.

So,

g = (v_{y_{f}} - v_{y_{i}} ) / t

9.8 =  14.6 - v_{y_{i}}) / 4

v_{y_{i}} = 24.6 m/s

v_{i} = \sqrt{v_{x_{i}}^{2}+v_{y_{i}}^{2} }

v_{i} = \sqrt{8.43^{2}+24.6^{2}}

v_{i} = 26 m/s

c)

cos β = v_{x_{i}} / v_{i}

β = cos⁻¹ (8.43 / 26)

β = 71.08°

3 0
3 years ago
sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o
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Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

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Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

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Answer:

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Energy, wavelength and frequency of a photon are connected using the expression:

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4 years ago
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