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Digiron [165]
3 years ago
6

A force that opposes motion in moving parts is friction. A. true b false

Physics
2 answers:
Cerrena [4.2K]3 years ago
8 0
A force that opposes motion in moving parts is friction. A
Julli [10]3 years ago
8 0
This is true. Friction is the resistance that one surface or object encounters when moving over another. The definition described above is worded differently but both definitions still mean the same thing.

Hope this helps!! ☺️
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A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 2.50 10
Katen [24]

Answer:

B = 6.18 10⁻⁶ T  

the magnetic field is in the negative direction of the y axis

Explanation:

The magnetic force is given by

         F = q v x B

as in the exercise indicate that the velocities perpendicular to the magnetic field,

         F = q v B

Newton's second law is

         F = m a

let's substitute

         q v B = m a

         B = m a / q v

let's calculate

         B = 9.1 10⁻³¹ 2.50 10¹³ / (1.6 10⁻¹⁹ 2.30 10⁷)

         B = 6.18 10⁻⁶ T

The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.

In the exercise indicate that the velocity is the z axis

the acceleration and therefore the force in the x axis

therefore the magnetic field is in the negative direction of the y axis

7 0
3 years ago
Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergo
oksian1 [2.3K]

Answer:

e). a = 0.066 m/s^2

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

\omega_f^2 - \omega_i^2 = 2\alpha \theta

now we have

\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})

\omega = 0.177 rad/s

now the centripetal acceleration of the point P is given as

a_c = \omega^2 R

a_c = (0.177)^2(2)

a_c = 0.063 m/s^2

tangential acceleration is given as

a_t = R\alpha

a_t = 2(0.01)

a_t = 0.02 m/s^2

now net acceleration is given as

a = \sqrt{a_t^2 + a_c^2}

a = \sqrt{0.02^2 + 0.063^2}

a = 0.066 m/s^2

8 0
3 years ago
How much force does it take to accelerate a 50.8 kg person at 3.50 m/s^2?
Vikki [24]

Apply Newton's second law to the person's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

m = 50.8kg, a = 3.50m/s²

Plug in and solve for F:

F = 50.8(3.50)

F = 178N

8 0
3 years ago
Why are some animals and plants can become pests when introduced into Australia
Yanka [14]
<span>Plants there are a lot. Australian plants are extremely hardy an have adapted to grow in conditions others have not. So when taken outside of Australia they do very, very well. Animals, less so. But there are a few very interesting cases.

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8 0
4 years ago
Kevin goes bowling. Whenever he bowls the ball, he transfers energy from his hand to the bowling ball. The amount of energy befo
g100num [7]

Answer:the answer is C

Explanation:

3 0
3 years ago
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