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stellarik [79]
3 years ago
8

A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r

adius 20.0 m. If an electron were moving at the same speed perpendicular to the same magnetic field, what would be the radius of the circle the electron would move in
Physics
1 answer:
Softa [21]3 years ago
5 0

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

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kvasek [131]

Based on the information given, it can be noted that the bat was either shorter or longer than what he expected.

From the information given, it was stated that Raghu wants a cricket bat to be made by a carpenter and he tells the carpenter that the length of the bat should be 7 hand spans.

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brainly.com/question/25292087

8 0
3 years ago
What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency
Anvisha [2.4K]

Answer:

1.04\times 10^{-7} T

Explanation:

IP  = Power of the bulb = 100 W

r  = distance from the bulb = 2.5 m

I = Intensity of light at the location

Intensity of the light at the location is given as

I = \frac{P}{4\pi r^{2}}

I = \frac{100}{4(3.14) (2.5)^{2}}

I = 1.28 W/m²

B_{o} = maximum magnetic field

Intensity is given as

I = \frac{B_{o}^{2}c}{2\mu _{o}}

1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}

B_{o} = 1.04\times 10^{-7} T

7 0
3 years ago
What is the force on a box that is being pushed to the right with 50n of force while, at the same time, being pushed to the left
elena-s [515]
I think the question meant to say net force on the box. Since force is a vector, the direction matters. 20N left is negated completely by the 50N right, which means the net force is 50N-20N to the right, 30N. 
7 0
3 years ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
Calculate the net force and the acceleration on the block
Darina [25.2K]

Answer:

Net Force = 10N

Acceleration = 2m/s^2

Explanation:

calculate the net force and the acceleration on the block

Net force on the block F = mass * acceleration

Net force acting in the positive direction = 4N + 6N = 10N

Mass = 5kg

According to newton's second law;

a = F/m

a = 10N/5

a = 2m/s^2

hence the acceleration on the block is 2m/s^2

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