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3 years ago

How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m

Physics

1 answer:

vovikov84 [41]3 years ago

8 0

<h2>Answer: 1000 J</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:

$1J=(1N)(1m)=Nm$

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (1)

When they are not parallel, both directions form an angle, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$ (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance $d$ to overcome the resistance of the weight of the box (its 200 N).

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work $W$ performed is the product of the force exerted to push the box $F=50N$ by the distance traveled $d$ . as shown in equation (1).

Hence:

$W=(50N)(20m)$

$W=1000Nm=1000J$ >>>>This is the work

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Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

MrMuchimi

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, $g=-9.81 m/s^2$ .

So, the final velocity is

$v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4$

$\Rightarrow v=-15.24$ m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

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3 years ago

Five metals that are more useful for electroplating

Radda [10]

5 meteals that are more useful for electroplating carbon,seth,

6 0

3 years ago

A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this

mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

7 0

3 years ago

2. A person applies a force of 66 N to a fridge as they push it across the length of a standard tennis court. So far today, the

Lubov Fominskaja [6]

Answer:

$P=39.2205\, watt$

$E=374.948 \,cal$

Explanation:

Given that:

force applied, $F=66\,N$

displacement, $s=23.77\,m$ (length of a tennis court)

time taken for pushing, t = 40 s

Since, work is given by:

$W=F.s$

$W=66\times 23.77$

$W=1568.82\,J$

Now, power is given as:

$P=\frac{W}{t}$

$P=\frac{1568.82}{40}$

$P=39.2205 \,watt$

Calories consumed is:

$E= 1568.82\times 0.239$

$E=374.948\, cal$

3 0

3 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Phoenix [80]

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is $1.60\times 10^{5}kg$ will require

$0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ$

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$

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3 years ago