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3 years ago

If the net torque is zero, what does this imply about the clockwise and counterclockwise torques.

Physics

1 answer:

lapo4ka [179]3 years ago

6 0

Answer:

Explanation:

When two forces acting on a line of action and they are equal in magnitude but opposite it direction, it forms a couple.

Torque is defined as the product of either force and the perpendicular distance between the two forces.

It is a vector quantity.

The net torque is zero, it means the anticlockwise torque is equal to the clockwise torque.

It means they balances each other.

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A proton traveling due west in a region that contains only a magnetic field experiences a vertically upward force (away from the

Vikki [24]

South

Explanation:

The magnetic force F on a point charge moving with a velocity v in the presence of a magnetic field B is given by

$\vec{\textbf{F}} = q\vec{\textbf{v}}\textbf{×}\vec{\textbf{B}}$

and according to the right-hand rule, an upward magnetic force on a proton moving westward is only possible if the magnetic field is directed southward.

3 0

3 years ago

Please help me with part b.

Lerok [7]

I'm sorry but I don't see anything that says part B
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8 0

3 years ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒ I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒ ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒ ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒ ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R

⇒ Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0

3 years ago

A gyroscope flywheel of radius 1.96 cm is accelerated from rest at 13.0 rad/s2 until its angular speed is 2270 rev/min. (a) What

nika2105 [10]

Tangential acceleration of a point on the rim of the flywheel during this spin-up process is 0.2548 m/s².

Tangential acceleration is defined as the rate of change of tangential velocity of the matter in the circular path.

Given,

Radius of flywheel (r) = 1.96 cm = 0.0196m

Angular acceleration (α)= 13.0 rad/s²

The tangential acceleration formula is at=rα

where, α is the angular acceleration, and r is the radius of the circle.

using the formula; at=rα = (13.0 rad/s²) (0.0196m) = 0.2548 m/s².

The tangential acceleration is 0.2548 m/s².

Learn more about the Tangential acceleration with the help of the following link:

brainly.com/question/15743294

#SPJ4

5 0

1 year ago

a 64kg skateboarder on a 2.0kg skateboard is on top of a ramp with a vertical height of 5.0 m what is the skateboarders maximum

WITCHER [35]

The formula for a kinetic energy KE of a falling body is
KE = mgh
where m = mass, g = acceleration due to gravity (9.8 m/s^2, constant), h = height.

The total mass of a skateboader and a skateboard is 64 + 2.0 = 66 kg.

Finally,
KE = 66*9.8*5.0 = 32340 J

7 0

3 years ago