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3 years ago

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

Physics

2 answers:

Sveta_85 [38]3 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]3 years ago

3 0

Answer:

Explanation:

Took it on edg

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The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca

4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t + 1/2 a t²

s = 1/2 a t² ----------- (2)

compare equation (1) and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0

3 years ago

A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block

Helga [31]

No friction present means: Ek = Ep

So Ek = mgh = 10 * 9.8 * 2 = 196 J

4 0

3 years ago

Read 2 more answers

A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220

disa [49]

First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, $\mu m g$ . For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
$ma=-\mu m g$
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
$a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2$

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
$2aS=v_f^2-v_i^2$
where $v_f=0$ is the final speed of the skier and $v_i=4.75 m/s$ is the initial speed. Substituting numbers, we find:
$S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m$