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pogonyaev
3 years ago
11

a crane does 9,500 J of work to lift a crate straight up using a force of 125 N. how high does the crane lift the crate?

Physics
1 answer:
Lina20 [59]3 years ago
7 0

Answer:

<em>The crane lifts the crate up to 76 m high</em>

Explanation:

<u>Work Done by a Force</u>

The work done by a force of magnitude F that displaces an object by a distance y is given by

W=F.y

We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.

The above equation can be solved to know the value of y in terms of the work and the force:

\displaystyle y=\frac{W}{F}

Plugging in the given values

\displaystyle y=\frac{9,500}{125}

y=76\ m

The crane lifts the crate up to 76 m high

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In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge press
GalinKa [24]

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

           

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = \sqrt{\frac{276000}{\rho }  - 43.675 }

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = \sqrt{\frac{276000}{1000} - 43.675}

           v₂ = 15.24 m / s

5 0
3 years ago
How can recycling materials lead to environmental sustainability?
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Oh my baby boy your back th anwsers is <span>A. Recycling materials help decrease the amount of new materials taken from the environment. </span>
6 0
4 years ago
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A radioactive sample has a count rate of 10,000 Bq. 24 days later the activity has fallen to 625 Bq. What is the half-life of th
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10000 Bq / 625 Pq = 16

Radioactivity has decreased by a factor of 16

2^4 = 16

So the sample has gone thru 4 half-lives

24 da / 4 = 6 da

6 da is the half-life

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What is the main reason why many nuclear power plants are located near bodies of water?
nasty-shy [4]
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4 0
4 years ago
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A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi
Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put  the value into the formula

-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2

u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}

u=-2.496\ m/s

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

v = u+gt

Put the value in the equation

-2.496=0-9.8\times t

t =\dfrac{2.496}{9.8}

t=0.254\ sec

The total time is

t'=t+1.95

t'=0.254+1.95

t'=2.204\ sec

We need to calculate the distance

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Put the value into the formula

s=0+\dfrac{1}{2}\times9.8\times(2.204)^2

s=23.80\ m

Hence, (a). The velocity of the object is -2.496 m/s.

(b).  The total distance of the object travels during the fall is 23.80 m.

4 0
3 years ago
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