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svetlana [45]
3 years ago
9

A cosmic ray electron moves at 6.5 x 106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is

1.24 10-5 T. What is the radius (in meters) of the circular path the electron follows? The charge of an electron is 1.6 x 10-19 C and its mass is 9.11 x 10-31m. Your answer should be a number with two decimal places, do not include the unit.
Physics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer: R=2.98 m

Explanation: In order to solve this answer we have to use the dynamic of circular movement actioned by the Lorentz force, this is given by:

Fm=qvB ( condiring that v and B are perpendicular)

By using the second Newton law fro a circular movement, we have:

Fm=m*a=m*v^2/R  where v and R are the speed and R of the circular movement electrons in the magnetic field.

then

qvB=m*v^2/R

Finally R=m*v/(qB)=9.11*10^-31*6.5*10^6/(1.6*10^-19*1.24*10^-5)=2.98

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Stella [2.4K]

Answer:

50 degree.

Explanation:

Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7

The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:

Tan Ø = Ay/Ax

Substitute Ay and Ax into the formula above.

Tan Ø = -4.7 / 5.6

Tan Ø = -0.839

Ø = tan^-1(-0. 839)

Ø = - 40 degree

Therefore, the angle between vector A and B positive direction of x-axis will be

90 - 40 = 50 degree.

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3 years ago
Write Newton's 3rd law. How would this law relate to a rocket ship taking off from the earth? Would this law affect the rocket s
irina1246 [14]

Answer:

Part A

Newton's 3rd law states that action and reaction are equal and opposite, mathematically, we have;

F_A = -F_B

Where;

F_A = The action force

F_B = The reaction force

Part B

The law indicates that the force with which a rocket ship uses in taking off from the Earth, F_A is equal in magnitude, and opposite in direction to the reaction force of the Earth to the motion of the rocket, (-)F_B

Part C

The law is a universal law, and it will also affect the rocket ship in space, as the force of the jet from the exhaust is directed towards Earth while in space, the rocket is propelled deeper into space

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Read 2 more answers
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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Atomic number is equal to the number of protons and electrons

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