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3 years ago

When a force is acting at axis of rotation,why torque is zero?

Physics

1 answer:

Vikki [24]3 years ago

4 0

Explanation:

see, torque=force × perpendicular distance

...that perpendicular distance is between axis of rotation and the point where force acts... so in above's case perpendicular distance is zero... so the torque is zero!

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A metal object is four feet away from a magnet. If you move the object two feet closer toward the

Temka [501]

The flux will be 4 times stronger ( due to the inverse square law )

8 0

4 years ago

A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w

garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

$E_s=\dfrac{1}{2}kx^2$

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

$E_P=mgx$

Equating these two energies,

$\dfrac{1}{2}kx^2=mgx$

$\dfrac{k}{m}=\dfrac{2g}{x}$ .............(1)

The expression for the oscillation frequency is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}$ (from equation (1))

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}$

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0

4 years ago

What is the power involved in lifting a 1.0-kg object 1.0 m in 1.0 s??

MAVERICK [17]

Mass = 1kg
Distance = 1m
Time = 1s
Force= Mass x Acceleration due to graviy
          = 1 x 9.8 = 9.8
Velocity = Distance / time
              = 1 / 1 =1m/s
Power = Force x velocity
           = 9.8 x 1 = 9.8 W

5 0

4 years ago

To test the performance of its tires, a car travels along a perfectly flat (no banking) circular track of radius 130 m. The car

marysya [2.9K]

Answer:

0.739

Explanation:

If we treat the four tire as single body then

W ( weight of the tyre ) = mass × acceleration due to gravity (g)

the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r

where v is speed 25.6 m/s and r is the radius of the circle

centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²

net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²

coefficient of static friction between the tires and the road = frictional force / force of normal

frictional force = m × net acceleration / m×g

where force of normal = weight of the body in opposite direction

coefficient of static friction = (7.2567 × m) / (9.81 × m)

coefficient of static friction = 0.739

4 0

3 years ago

Part e a small toy cart equipped with a spring bumper rolls toward a wall with a speed of v. the cart rebounds from the wall, wi

jeka94

Answer:

Δp = -2 p₀

Explanation:

The momentum is defined by

p = m v

In this case we write the initial and final momentum, we take as positive the direction towards the wall.

p₀ = m v

p_f = m (-v)

the negative sign is because the car is bouncing off the wall

the change of the moment is

Δp = p_f - p₀

Δp = - m v - m v

Δp = -2 mv

Δp = -2 p₀

we see that the change of moment is twice the moment, in the attachment we can see the vectors of these changes and the sign indicates the direction of the change at the moment

5 0

3 years ago