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Gelneren [198K]
2 years ago
8

A particle is located on the x axis at x = 2.0 m from the origin. A force of 25 N, directed 30° above the x axis in the x-y plan

e, acts on the particle. What is the torque about the origin on the particle?
Physics
1 answer:
nata0808 [166]2 years ago
7 0

Answer:

25 Nm

Explanation:

Parameters given

Distance of the particle from the origin, r = 2.0 m

Force acting on particle, F = 25 N

Angle of force, θ = 30°

The torque acting on a particle is given as:

τ = r * F * sinθ

Where r = radius of axis of rotation. This is the same as the position of the particle on the x axis.

Therefore, the torque will be:

τ = 2 * 25 * sin30

τ = 25 Nm

The torque acting on the particle is 25 Nm.

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Suppose that air resistance cannot be ignored. For the position at which the person has jumped from the platform and the cord re
Kamila [148]

Answer:

The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.

Explanation:

In case, where air resistance is being ignored the stretch cord will stretch as it normally does.              

  • Air resistance is a force that any object experiences as a result of its motion through the air.  

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2 years ago
Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between
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<span>Two objects move toward each other because of gravitational attraction. As the objects get closer and closer, the force between them increases. </span>
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2 years ago
Read 2 more answers
A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0
2 years ago
If the average speed of an orbiting space shuttle is 27 800 km/h, determine the time required for it to circle Earth. Assume tha
balu736 [363]

Answer:

 t = 1.51 hours

Explanation:

given,

Speed of space shuttle. v = 27800 Km/h

Radius of earth, R = 6380 Km

height of shuttle above earth, h = 320 Km

Total radius of the shuttle orbit

r' = R + h

r' = 6380 + 320

r' = 6700 Km

distance, d = 2 π r

   d = 2 π x 6700

time = \dfrac{distance}{speed}

time = \dfrac{2\pi\times 6700}{27800}

 t = 1.51 hours

Time require by the shuttle to circle the earth is equal to 1.51 hr.

7 0
2 years ago
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
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