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dlinn [17]
1 year ago
6

A cosmic-ray electron moves at perpendicular to earth’s magnetic field at an altitude where the field strength is. what is the r

adius of the circular path the electron follows?
Physics
1 answer:
Harlamova29_29 [7]1 year ago
6 0

4.266 m is the radius of the circular path the electron follows.

Given

Speed of electron (v) = 7.5 × 10⁶ m/s

Earth's Magnetic Field (B) = 1 × 10⁻⁵ T

We already know that

Mass of electron (m) = 9.1 × 10⁻³¹ kg

Charge on electron (q) = 1.6 × 10⁻¹⁹ C

According to the formula

Radius of circular path(r) = mass on electron × speed/ Charge × Magnetic field

Radius of circular path(r) = m × v/q × B

Put the values into the formula

r = 9.1 × 10⁻³¹ × 7.5 × 10⁶/ 1.6 × 10⁻¹⁹ × 10⁻⁵

On solving, we get

r = 4.266 m

Hence, 4.266 m is the radius of the circular path the electron follows.

Learn more about magnetic field here brainly.com/question/26257705

#SPJ4

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If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times t
Tasya [4]

Answer:

F = K Q1 Q2 / R^2       force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

8 0
2 years ago
What is electrical power and how is it measured?
Sliva [168]

Answer:

power is measured in watts and its the messure of work over time.

Explanation:

8 0
3 years ago
Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each ar
liberstina [14]

Answer:

 ∑ τ =0,  L₀ = L_{f}

Explanation:

In a circular turning movement, when the arms are extended and then contracted in two possibilities:

- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.

    In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.

- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted

 

In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant

          ∑ τ =0

                L₀ = L_{f}

              I₀ w₀ = I w

4 0
3 years ago
A 60 g ball of clay is thrown horizontally at 40 m/s toward a 1.5 kg block sitting at rest on a frictionless surface. the clay h
Bingel [31]
The solution for this problem is:
Let u denote speed. 

Equating momentum before and after collision: 
= 0.060 * 40 = (1.5 + 0.060) u 
= 2.4 = 1.56 u
= 2.4 / 1.56 = 1.56 u / 1.56
= 1.6 m / s is the answer for this question. This is the speed after the collision.
7 0
3 years ago
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
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