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3241004551 [841]
2 years ago
14

If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?

Physics
2 answers:
Sveta_85 [38]2 years ago
6 0

Answer:

The velocity of charge is 1.1×10⁴

Option(D)

Explanation:

F = q(v×B)

3.5×10-²= (8.4×10-⁴)*(v)*(6.7×10-³)*sin35

v = (3.5)/(8.4×6.7×0.57) × 10⁵

v = 3.5/32.08 × 10⁵

v = 0.109 × 10⁵

v ≈ 1.1 × 10⁴ m/s

zaharov [31]2 years ago
3 0

Answer:

D

Explanation:

Took it on edg

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When fuel and air are compressed in the compression stroke, ...... a. each molecule of fuel is heated to its flash point b. each
alexira [117]

None of the choices is an appropriate response.

There's no such thing as the temperature of a molecule.  Temperature and
pressure are both outside-world manifestations of the energy the molecules
have.  But on the molecular level, what it is is the kinetic energy with which
they're all scurrying around.

When the fuel/air mixture is compressed during the compression stroke,
the temperature is raised to the flash point of the mixture.  The work done
during the compression pumps energy into the molecules, their kinetic
energy increases, and they begin scurrying around fast enough so that
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7 0
3 years ago
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular
Evgesh-ka [11]

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

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

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}

<u>α = - 1930.2 rad/s²</u>

<u>negative sign shows deceleration</u>

5 0
3 years ago
A toy car pushed across the floor is observed to slow down and stop based on Newton’s first law of motion what is the best concl
Maksim231197 [3]

An object in motion eventually stops because of friction.

<h3>What is Newton’s first law of motion?</h3>

Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or continue its moving in a straight line at constant speed unless force is applied on it.

So we can conclude that An object in motion eventually stops is the correct answer.

Learn more about law here: brainly.com/question/820417

3 0
2 years ago
A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar
sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

v^2-u^2 = 2gd

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

y_1 = h - \frac{1}{2}gt^2

While the vertical position of the ball thrown upward is

y_2 = ut - \frac{1}{2}gt^2

The two balls meet when

y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m

So the distance below the top of the cliff is

d=24.0 - 18.0 = 6.0 m

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3 years ago
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