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LuckyWell [14K]
3 years ago
13

a loaded sack of total mass is 1000 gramme falls down from the floor of a lorry 200cm high, calculate the workdone by the gravit

y of the loads​
Physics
1 answer:
Helen [10]3 years ago
8 0

Answer:

W = 20 J

Explanation:

Given that,

The mass of a loaded sack, m = 1000 g = 1 kg

It falls down from the floor of a lorry 200 cm high, h = 2 m

We need to find the work done by the gravity. The work done by an object under the action of gravity is given by :

W = mgh

Substitute all the values,

W = 1 × 10 × 2

= 20 J

Hence, the required work done by gravity is equal to 20 J.

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100 point <br> good luck answering this<br> 2+2-4+8+-5+???
Darya [45]
The answer to that is 3
7 0
3 years ago
PLEASE ANSWER FAST In which of the following situations is the greatest amount of work accomplished? 1. A boy lifts a 2-newton b
GuDViN [60]

Explanation:

Work done is given by the product of force and displacement.

Case 1,

1. A boy lifts a 2-newton box 0.8 meters.

W = 2 N × 0.8 m = 1.6 J

2. A boy lifts a 5-newton box 0.8 meters.

W = 5 N × 0.8 m = 4 J

3. A boy lifts a 8-newton box 0.2 meters.

W = 8 N × 0.2 m = 1.6 J

4. A boy lifts a 10-newton box 0.2 meters.

W = 10 N × 0.2 m = 2 J

Out of the four options, in option (2) ''A boy lifts a 5-newton box 0.8 meters'', the work done is 4 J. Hence, the greatest work done is 4 J.

3 0
3 years ago
The _____ deflects these winds to the right in the northern hemisphere and to the Southern Hemisphere
almond37 [142]

Answer:

Coriolis Effect

Explanation:

The Coriolis effect is responsible for the deflection of winds to the right in the Northern hemisphere and to the right in the Southern hemisphere. It is an effect that occurs because of the rotation of the earth around its axis.

The implication of this is that in areas of low pressure in the Northern hemisphere, winds tend to blow in anticlockwise direction, and in areas of high pressure, it blows in a clockwise direction. The opposite of this happens in the Southern hemisphere.

6 0
3 years ago
Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.
aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

 

6 = (20 / 32.2 + 50 / 32.2)a

 

2.173a = 6

 

A = 2.76 ft/s^2

 

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

 

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0
3 years ago
For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro
Mariulka [41]

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

v = u + gt

Where v is the final velocity

u is the initial velocity

g is acceleration due to gravity (take g = 9.8 m/s²)

and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

Also, from one of the equations of kinematic for free fall motions,

h = ut + 1/2(gt²)

u = 0 m/s

From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore

h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

5 0
3 years ago
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