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Ierofanga [76]
3 years ago
10

The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe

d to the top of the building
it would help if work was shown :,)
Physics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

 Work done:

             Work done  = Weight x height  = 700 x 828

        Work done  = 579600J

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A 46 g domino slides down a 30 degrees incline at a constant speed. What is the coefficient of friction?
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Answer:

40

Explanation:

30

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If you are in an elevator that speed up then the apparent weight is
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Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

W_{net} = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

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3 years ago
2. An elephant puts a force of 30,000 N on its four feet, which have a
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3 years ago
Definition of complex machine
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Read 2 more answers
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
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