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2 years ago

What work is done by a forklift raising a 574 kg crate to a height of 1.3m in a time of 2.0 seconds?

Physics

1 answer:

Zinaida [17]2 years ago

4 0

Answer: $7312.76\ J$

Explanation:

Given

Mass of crate is m=574 kg

The crate is raised up to a height of 1.3 m

time taken t=2 s

Work done to raise the crate is equal to the change in its potential energy

$\Rightarrow \text{Work done W=}mgh\\\Rightarrow W=574\times 9.8\times 1.3\\\Rightarrow W=7312.76\ J$

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a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will

Lisa [10]

Answer:

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Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

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3 years ago

A running back with a mass of 70 kg travels down the field with a velocity of 5.0 m s . Calculate the kinetic energy of the foot

jarptica [38.1K]

K.E = 1/2 mv²

= 1/2 (70kg) (5.0ms)²

= 1 ˣ ( 1750) / 2

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3 years ago

Read 2 more answers

Which example would most likely decrease friction?

andreev551 [17]

Answer:

using cleats other than gym shoes

Explanation:

the cleats hwve spikes and help gain speed

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3 years ago

8TH GRADE SCIENCE I NEED NOW DO NOT SKIP

yan [13]

Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

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7 0

3 years ago

A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

Veronika [31]

Answer:

vf = 3.27[m/s]

Explanation:

In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.

With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.

With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.

Then we equalize the two stress equations and we can clear the acceleration.

a = 3.58 [m/s^2]

As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.

$x = \frac{1}{2} *a*t^{2}\\1.5 = \frac{1}{2}*(3.58) *t^{2} \\t = 0.91 [s]$

And the speed can be calculated as follows:

$v_{f}=v_{o}+a*t\\v_{f}=0+(3.58*0.915)\\v_{f}= 3.27[m/s]$

7 0

3 years ago