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Degger [83]
3 years ago
11

If a negatively charged particle enters a region of uniform magnetic field which is perpendicular to the particle's velocity, wi

ll the kinetic energy of the particle increase, decrease, or stay the same?
Physics
1 answer:
lys-0071 [83]3 years ago
3 0

Answer:

Same

Explanation:

While moving through a magnetic field in a direction perpendicular to a B-field, a continuous force experienced by a charged particle. If this magnetic field remains uniform, the force exerted also remains same and hence the velocity with which the particle is moving remains same. However, the particle is forced to move on a curved path until it forms a complete circle.

Hence, the kinetic energy remains the same because the speed is same

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The Prime Meridian

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3 years ago
A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10
bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

m_1 = Mass of sphere = 10000 kg

m_2 = Mass of rod = 10 kg

r = Radius of sphere = 2 m

l = Length of antenna = 3 m

Angular speed

\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s

Angular momentum is given by

L=I\omega

Moment of inertia of the satellite is

I_s=\frac{2}{5}m_1r^2

Moment of antenna of the satellite is

I_a=\frac{1}{3}m_2l^2

The angular momentum of the system is

L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s

The angular momentum of the satellite is 605447.7066 kgm²/s

5 0
3 years ago
A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co
exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0
3 years ago
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