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ArbitrLikvidat [17]

3 years ago

9

Can an electron at rest in a magnetic field be set into motion by the magnetic field? What if it were at rest in an electric fie

ld?

1 answer:

Burka [1]3 years ago

7 0

Answer:

By a magnetic field: no

By an electric field: yes

Explanation:

The force exerted by a magnetic field on an electron is

$F=qvB sin \theta$

where

q is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

$\theta$ is the angle between the direction of v and B

As we see from the formula, if the electron is at rest, then v = 0, and therefore the force is also zero: F = 0. Therefore, the magnetic field cannot set the electron into motion.

On the other hand, the force exerted on an electron by an electric field does not depend on the speed:

$F=qE$

where E is the intensity of the electric field

Therefore, the electric force acts also when the electron is at rest, so it is able to set the electron into motion.

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It always takes _______ to change the motion of an object.

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Answer:

It always takes energy to change the motion of an object.

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3 years ago

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A 25.0kg girl pushes a 50.0kg boy with a force of 100.0N. What is the acceleration of the girl?

MA_775_DIABLO [31]

Answer:

im pretty sure it should be 50.0

6 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

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3 0

4 years ago

incline plane is given length 12m,load 600 newton,effort 200 Newton, Height 3 metre find its velocity ratio and mechanical advan

Salsk061 [2.6K]

Answer:

i. The velocity ratio of the plane is 4.

ii. The mechanical advantage of the plane is 3.

Explanation:

i. The velocity ratio (VR) of an inclined plane is ratio of its length to the height. It is given as;

VR = $\frac{length of the plane}{height}$ = $\frac{l}{h}$

Given: l = 12 m, L = 600 N, E = 200 N, h = 3 m.

So that,

VR = $\frac{12}{3}$

= 4

The velocity ratio of the plane is 4.

ii. Mechanical advantage (MA) expresses the relationship between the load overcome to effort applied.

MA = $\frac{Load}{Effort}$ = $\frac{L}{E}$

= $\frac{600}{200}$

= 3

The mechanical advantage of the plane is 3.

Therefore, the velocity ratio of the inclined plane is 4, and its mechanical advantage is 3.

7 0

3 years ago

g A spherical container of inner diameter 0.9 meters contains nuclear waste that generates heat at the rate of 872 W/m3. Estimat

zhannawk [14.2K]

Answer: The total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.

Explanation:

Given: Inner diameter = 0.9 m

q = 872 $W/m^{3}$

Now, radii is calculated as follows.

$r = \frac{diameter}{2}\\= \frac{0.9}{2}\\= 0.45 m$

Hence, the rate of heat transfer is as follows.

$Q = q \times V$

where,

V = volume of sphere = $\frac{4}{3} \pi r^{3}$

Substitute the values into above formula as follows.

$Q = q \times \frac{4}{3} \pi r^{3}\\= 872 W/m^{3} \times \frac{4}{3} \times 3.14 \times (0.45 m)^{3}\\= 332.67 W$

Thus, we can conclude that the total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.

3 0

3 years ago