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hammer [34]
4 years ago
10

Usually the force of gravity on electrons is neglected. To see why, we can compare the force of the Earth’s gravity on an electr

on with the force exerted on the electron by an electric field of magnitude of 40000 V/m (a relatively small field). What is the force exerted on the electron by an electric field of magnitude of 40000 V/m? The acceleration of gravity is 9.8 m/s 2 , the mass of an electron is 9.10939 × 10−31 kg, and the elementary charge 1.602 × 10−19 C. Answer in units of N.
Physics
1 answer:
miv72 [106K]4 years ago
6 0

Answer:

6.4\cdot 10^{-15} N

Explanation:

The electric force exerted on the electron is given by:

F=qE

where

q=1.6\cdot 10^{-19}C is the magnitude of the electron charge

E = 40000 V/m is the electric field

Substituting,

F=(1.6\cdot 10^{-19} C)(40000 V/m)=6.4\cdot 10^{-15} N

By comparison, the gravitational force exerted on the electron is:

F=mg

where

m=9.10939\cdot 10^{-31} kg is the mass of the electron

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting,

F=(9.10939\cdot 10^{-31} kg)(9.8 m/s^2)=8.93\cdot 10^{-30}N

You might be interested in
Some cities now implement signal lights designed to specifically apply to _____ rather than motorized vehicles.
dmitriy555 [2]

The correct answer is C; Bicycles.

Further Explanation:

In major cities, in the United States, have implemented signal lights specifically designed for bicycle riders. The riders also have their own designated bike lanes in many large cities. Drivers in vehicles, are to give the right of way to people on bicycles.

Bicycle riders are to follow the same laws and laws specifically for the riders or they can face fines and tickets.

Learn more about bicycle laws at brainly.com/question/8934107

#LearnwithBrainly

7 0
3 years ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
How much power does it take to lift a 30.0 n box 10.0 m high in 5.00 s, if you must apply a 62n force to lift the box?
Illusion [34]
Power is defined as the rate at which the body is doing work:
P=\frac{W}{t}
Work is defined as displacement done by the force times that displacement:
W=F\cdot h
We know that we need 62N to move the box, so when we apply this force along the path of 10m we have done:
W=62N\cdot10m=620J
of work.
Now we just divide that by 5s to get how much power is required:
P=\frac{620J}{5s}=124W
5 0
3 years ago
A 60-watt light bulb has a voltage of 120 volts applied across it and a current of 0.5 amperes flows through the bulb. What is t
Naddik [55]

Answer: 240ohms

Explanation: P=IV

P=60watts

I=0.5A

V=120volts

From ohms law

V=IR

R=V/I

R=120/0.5

R= 240ohms

7 0
3 years ago
A car manufacturer wants to change its car’s design to increase the car’s acceleration. Which changes should the engineers consi
likoan [24]

Answer:

The car manufacturers could increase bore of the cylinders, place the engine in the center or back of the car, add 1 to 2 turbochargers, and lower the center of gravity of the vehicle to increase traction.

Explanation:

Turbochargers would be recommended because they significantly increase both the torque of the engine as well as the amount of horses powering the car while also increasing original efficiency both with and without the additional power. Weight adjustment allows for lightweight vehicles with good traction. This is important to both keep control of the car under acceleration, but it also makes the vehicle more efficient due to the now sheddable unnecessary weight. A more obvious approach would be to increase the base horsepower and torque of the engine by increasing the bore of the cylinders and the weight of the pistons. This acts as an inertial lever, because the extra piston weight will drag the crankshaft faster. This could also be achieved by taking away piston weight, but this could be catastrophic should a piston slip.

4 0
3 years ago
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