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2 years ago

10

A forklift does 450 J of work to lift a 150 N hot tub. How high is the hot tub lifted? Round your answer to the nearest whole nu

mber.
The hot tub is lifted BLANK m.

2 answers:

OLEGan [10]2 years ago

7 0

Answer:

The hot tub is lifted 3 m.

vivado [14]2 years ago

5 0

There are already 2 givens, hence we can do direct substitution to get the answer. To make the process simpler, derive the distance formula from the Work formula.

Work = Force x Distance

Distance = $\frac{Work}{Force}$

Work is 450J while the force is 150N hot tub

To get the proper units, get the equivalent of Joule to eliminate newton. A joule Is equal to 1 N-m

Distance = $\frac{450N-m}{150N}$
Distance = 3m

Hence, the hot tub is lifted 3 meters.

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2 years ago

Read 2 more answers

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

Anna007 [38]

Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

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$490=360+W_{f}$

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