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Veseljchak [2.6K]
2 years ago
6

A 2.0-mm-diameter glass sphere has a charge of 1.0 nC. What speed does an electron need to orbit the sphere 1.0 mm above the sur

face
Physics
1 answer:
Lorico [155]2 years ago
3 0

The electron would need to attain a speed of at least 2.813*10^7 m/s

Data;

  • diameter of sphere = 2.0mm
  • charge = 1.0 nC

The distance of the electron from the center of the sphere  = (1 + 1)mm = 2mm.

2mm = 2.0*10^-3m

<h3>Centrifugal Force and Coulomb's law</h3>

Using centrifugal force of attraction and coulomb's law, we can determine the speed which the electron needs.

\frac{mV^2}{R}=\frac{Kq_1q_2}{r^2}\\ K = \frac{1}{4\pi \epsilon } \\ V^2 = \frac{q_1R\\}{4\pi \epsilon \ m r} \\v = \sqrt{\frac{1.0*10^-^9*1.602*10^-^1^9}{4\pi \epsilon *9.1*10^-^3^1*2.0*10^_3} } \\

Solving the above, we would have

v = 2.813*10^7m/s

The electron would need to attain a speed of at least 2.813*10^7 m/s

Learn more on centrifugal and coulomb's law here;

brainly.com/question/24743340

brainly.com/question/10220774

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A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
alisha [4.7K]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

4 0
3 years ago
A lab cart is loaded with different masses and moved at various velocities
Sedaia [141]
A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

1.0m/s → 4kg
6 0
3 years ago
Discuss impact of Newton's second law on roads​
ruslelena [56]

Answer:

The second law: When a force is applied to a car, the change in motion is proportional to the force divided by the mass of the car. This law is expressed by the famous equation F = ma, where F is a force, m is the mass of the car, and a is the acceleration, or change in motion, of the car

4 0
2 years ago
Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3
Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

k_{1}=20\ N/m

k_{2}=30\ N/m

k_{3}=15\ N/m

k_{4}=20\ N/m

k_{5}=35\ N/m

According to figure,

k_{2} and k_{3} is in series

We need to calculate the equivalent

Using formula for series

\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}

k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}

Put the value into the formula

k=\dfrac{30\times15}{30+15}

k=10\ N/m

k and k_{4} is in parallel

We need to calculate the k'

Using formula for parallel

k'=k+k_{4}

Put the value into the formula

k'=10+20

k'=30\ N/m

k_{1},k' and k_{5} is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}

Put the value into the formula

k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}

k_{eq}=8.93\ N/m

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0
3 years ago
Which of the following accurately shows how to calculate the weight of a 20kg object
forsale [732]
The answer to your question is "20kgx9.8m/s" because weight is the force an object is exerting on another object, and the formula used to calculate force is <em>Force = Mass * Acceleration</em>.
4 0
3 years ago
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