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NeX [460]
2 years ago
6

Q12. How much force is applied on the body when 150 joule of work

Physics
1 answer:
ra1l [238]2 years ago
4 0

Force = Work/distance

Force = 150/10

          =  15 Newtons

Force = 15 Newtons

Therefore, 15 newtons of force is applied to the body when 150 joules of work

is done in displacing the body through a distance of 10m in the  direction of the force.

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Positive electric charge and found inside the nucleus
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Consider this situation: Four ropes, each attached to the end
faust18 [17]

The forces acting on the elevator are:

Gravity force

Tension force

Air resistance

Explanation:

Let's go through each of the forces listed and see which ones are acting on the elevator.

  • Normal force: NO. The normal force is a force exerted by a surface whenever there is another object "pushing" on it. For instance, when a box is at rest on a table, the box is "pushing" on the table (due to its weight), and the table "pushes back" on the box, upward, in order to balance its weight: this is the normal force. In this case, the elevator is lifted, so it is not pushing on anything, therefore there is no normal force.
  • Gravity force: YES. The force of gravity acts on every object located in the gravitational field of the Earth; it pulls downward, and its magnitude is mg, where m is the mass of the object and g is the acceleration of gravity.
  • Applied force: NO. Here there is no applied force, since there is nobody "pushing" or "pulling" the elevator.
  • Friction force: NO. As we are considering the forces on the elevator, and the elevator is not sliding against any surfaces, there is no force of friction. (The force of friction acts whenever there are two surfaces sliding against each other, which is not the case here)
  • Tension force: YES. The tension force is the force exerted by a rope or a string when pulling an object. In this case, there are four ropes pulling the elevator, therefore there are 4 forces of tension acting on the elevator, upward.
  • Air resistance: YES. As the elevator is moving through the air, the interaction between the molecules of air with the surface of the elevator produces a force (called air resistance) that "resists" the motion of the elevator, therefore pushing downward. However, the magnitude of this force is negligible in this case.

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

5 0
3 years ago
If I want to calculate the normal force of an object on two legs but the weight of the object is not uniform throughout, how do
RoseWind [281]

Use the right equation. To calculate the normal force of an object at an angle, you need to use the formula: N = m * g * cos (x) For this equation, N refers to the normal force, m refers to the object's mass, g refers to the acceleration of gravity, and x refers to the angle of incline.

6 0
2 years ago
As time goes on, the ENTROPY in a closed system should increase. This is because of which Law?
olga_2 [115]

Answer:

The answer is D

The second law of thermodynamics

5 0
3 years ago
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br
sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

6 0
3 years ago
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