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scoray [572]
2 years ago
5

A crate is lifted vertically 1.5 m and then held at rest. The crate has weight 100 N (i.e., it is reese (enr647) – HS OnRamps 04

: Intro to Work and Energy – mcdaniel – (34472020) 2 supported by an upward force of 100 N). How much work is being done to hold the crate 1.5 m above the ground in this way?
Physics
2 answers:
SOVA2 [1]2 years ago
8 0

Answer:150Joules

Explanation:

Work is said to be done if a force is applied to cause a body to move through a distance.

Work done = Force × distance

Given force = 100N

Distance = 1.5m

Work done = 100 × 1.5

Work done = 150Joules.

icang [17]2 years ago
5 0

Answer:

W = 0 J

Explanation:

It is given that,

Weight of the crate, W = 100 N

Distance moved by the crate, d = 1.5 m

Let W is the work done to hold the crate 1.5 m above the ground in this way. It is given by the product of force and the displacement. Its formula is given by :

W=F\times d\times cos\theta

Here, \theta=90^{\circ} as it is lifted vertically

W=100\ N\times 1.5\ m\ cos(90)      

W = 0

So, the work done to hold the crate is 0. Hence, this is the required solution.

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3. A 75kg man sits at one end of a uniform seesaw pivoted at its center, and his 24kg son sits at the
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The wife have to sit at 0.46 L from the middle point of the seesaw.

Explanation:

We need to make a sketch of the seesaw and the loads acting over it.

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A logical intuition will give us the idea that the mother will be on the side of her son to make the balance.

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2 years ago
An antibaryon composed of two antiup quarks
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(2) −1 e

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The antiparticle of down quark is antidown quark and has charge +\frac{1}{3}e charge.

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2 years ago
The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f
zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

                 584 = Ha + Lg

an also this mean that torques acting on the body is balanced

         So,   0.410 Ha  = 0.840 Lg

    Making Lg the subject of formula in the equation above we

   Lg = 0.4881 Ha

 Considering the first equation and replacing Lg with this recent equation we have

                      584 = Ha + 0.4881 Ha

          Therefore Ha = 392.44 N

This value obtained is  for both hands for each hand we divide by 2

Therefore we have for each hand = 392.44/2 =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

             Lg = 0.4881 ×  392.44

                  = 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

                  191.22/2 = 95.8 N.

8 0
2 years ago
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Lisa [10]
Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J
3 0
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