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4 years ago

An electric motor and a single-fixed pulley work together to lift a 500 kg crate 50.0 m. How much work was done?

Physics

2 answers:

tamaranim1 [39]4 years ago

8 0

Answer : Work done, W = 245,000 J

Explanation :

It is given that,

Mass of the system, m = 500 kg

Distance, d = 50 m

According to the definition of work done, $W=force\times displacement$

So, work done to lift the system of electric motor and a pulley is :

$W=mg\times h$

$W=500\ kg\times 9.8\ m/s^2\times 50\ m$

$W=245,000\ J$

Hence, this is the required solution.

tester [92]4 years ago

3 0

Answer: The work done is 245000 J.

Explanation:

The formula for the force due to the weight of the object is as follows;

$F=mg$

Here, m is the mass of the object and g is the acceleration due to gravity.

Put m=500 kg and g=9.8 meter per second.

$F=(500)(9.8)$

$F=4900 N$

The formula for the work done is as follows;

$W=Fs$

Here, F is the force and s is the displacement.

Put F=4900 N and s=50.0 m.

$W=(4900)(50)$

$W=245000 J$

Therefore, the work done is 245000 J.

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Sarah's mother gets a flat tire on her car while driving Sarah to school. They use a jack to change the tire. It exerts a force

3241004551 [841]

Answer:

1250 J

Explanation:

Work is said to be done when a force causes an object to move over a distance. The amount of work done (W) is calculated by multiplying the force by the distance traveled.

That is;

W = F × d

Where;

W = work done (J or N/m)

F = force (N)

d = distance (m)

Based on the information provided in this question, F = 5000N, d = 0.25m

Hence;

W = F × d

W = 5000 × 0.25

W = 1250J

Therefore, 1250Joules of work is done by the jack.

5 0

3 years ago

If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

Nat2105 [25]

A ) decrease.
B ) increase.
C ) increase, then decrease.
D ) not change.

The answer is A) decrease

Take pushing a box, for example-- You push your hardest then give out, still trying to push the box. You are doing less work than what you have started with!

( Mind marking me for branliest? ; ) )

4 0

3 years ago

A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

3 years ago

A small sphere with mass mcarries a positive chargeqand is attached to one end of a silk fiber of lengthL. The other end of the

Aleksandr-060686 [28]

Answer:

(a): The magnitude of the electric force on the small sphere = $\dfrac{q\sigma}{2\epsilon_o}.$

(b): Shown below.

Explanation:

<u>Given:</u>

m = mass of the small sphere.
q = charge on the small sphere.
L = length of the silk fiber.
$\sigma$ = surface charge density of the large vertical insulating sheet.

When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:

$\rm E = \dfrac{\sigma}{2\epsilon_o}.$

<em>where,</em>

$\epsilon_o$ is the electrical permittivity of the free space.

The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:

$\rm F_e=qE.$

Thus, the magnitude of the electric force on the small sphere is given by

$\rm F_e = q\times \dfrac{\sigma }{2\epsilon_o}=\dfrac{q\sigma}{2\epsilon_o}.$

The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.

When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.

According to the fig.,

$\rm \tan \theta = \dfrac{F_e}{W}.$

<em>where,</em>

$\rm F_e$ = electric force on the sphere, acting along left.
$\rm W$ = weight of the sphere, acting vertically downwards.

<em />

$\rm F_e = \dfrac{q\sigma}{2\epsilon_o}\\\\W=mg\\\\Therefore,\\\\\tan\theta = \dfrac{\dfrac{q\sigma}{2\epsilon_o}}{mg}=\dfrac{q\sigma}{2mg\epsilon_o}.\\\Rightarrow \theta=\tan^{-1}\left ( \dfrac{q\sigma}{2mg\epsilon_o}\right ) .$

g is the acceleration due to gravity.

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3 years ago

Running, swimming, and biking are what type of exercise aerobic or anaerobic

sesenic [268]

They are all cardiovascular exercises.

7 0

3 years ago

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