1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Setler [38]
2 years ago
11

Assuming a mass of 0.04 g, what is the power exerted by the electrical forces on a fragment of fiber if the fragment has a veloc

ity of 200 mm/s, 10 ms after ejection? For this problem, I used the formula P= Force * velocity = mass* (velocity/time)*velocity = (mass * velocity^2) /time. However I got the problem wrong. In the answer, they used the formula P= W/t = 1/2* (mass*velocity^2)/time. I am confused.When do you use P = Force * velocity = (mass * velocity^2) /time and when do you use P= W/t = 1/2* (mass*velocity^2)/time?
Physics
1 answer:
schepotkina [342]2 years ago
8 0

The power exerted by the electrical forces on a fragment of fiber is 8.0× 10^-5 Watts

<h3>Power and Energy or Work done</h3>

Power is the rate of doing work.

  • Power = work done / time

Energy is the ability to do work.

Work done and energy are used interchangeably.

  • Work done = Force × distance

In the example above, the energy possessed by the mass is kinetic energy.

Kinetic energy is the energy a body has by virtue of its motion.

  • Kinetic energy, KE = 1/2 mv^2

Calculating kinetic energy of the body:

mass of the body = 0.04 g = 4.0 × 10^-5 kg

velocity of the body = 200mm/s = 0.2 m/s

Kinetic energy of the body = 1/2 × 4.0 × 10^-5 × 0.2^2

Kinetic energy of the body = 8.0× 10^-7

Thus, work done = 8.0× 10^-7 J

Calculating power exerted:

time taken = 10 ms = 0.01 s

Power = work done/ time

Power exerted = 8.0× 10^-7/0.01

Power exerted = 8.0× 10^-5 Watts

Therefore, the power exerted by the electrical forces on a fragment of fiber is 8.0× 10^-5 Watts

Learn more about about electrical energy and power at: brainly.com/question/60890

You might be interested in
A 2.0-kg mass is oscillating about the origin at 24 rad/s. The amplitude of the oscillations is 0.040 m. At what position is the
Darya [45]

Answer:

0.0327 m

Explanation:

m = 2 kg

ω = 24 rad/s

A = 0.040 m

Let at position y, the potential energy is twice the kinetic energy.

The potential energy is given by

U = 1/2 m x ω² x y²

The kinetic energy is given by

K = 1/2 m x ω² x (A² - y²)

Equate both the energies as according to the question

1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)

y² = 2 A² - 2 y²

3y² = 2A²

y² = 2/3 A²

y = 0.82 A = 0.82 x 0.040 = 0.0327 m

4 0
3 years ago
Please help me
gtnhenbr [62]

Answer:

Catch these points g thanks

Explanation:

0.1

4 0
2 years ago
Read 2 more answers
A tradesman sharpens a knife by pushing it with a constant force against the rim of a grindstone. The 30-cm-diameter stone is sp
Lorico [155]

Answer:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2

b) Assume the grind stone is a solid disk, its moment of inertia is

I = mR^2/2

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

I = 28*0.15^2/2 = 0.315 kgm^2

So the friction torque is

T_f = I\alpha = 0.315*0.21 = 0.06615 Nm

The friction force is

F_f = T_f/R = 0.06615 / 0.15 = 0.441 N

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

N = F_f/\mu = 0.441/0.2 = 2.205 N

7 0
2 years ago
Which of the following represents thermal energy transfer through radiation
Bogdan [553]

C. since the the heat from the heater is going to the child in <u>waves</u>, it’s<u> radiating </u>

3 0
3 years ago
A batter hits a fly ball which leaves the bat 0.89 m above the ground at an angle of 62 ∘ with an initial speed of 29 m/s headin
KatRina [158]

consider the motion in Y-direction

v₀ = initial velocity = 29 Sin62 = 25.6 m/s

a = acceleration = - 9.8 m/s²

t = time of travel

Y = vertical displacement = - 0.89 m

using the equation

Y = v₀ t + (0.5) a t²

- 0.89 = (25.6) t + (0.5) (- 9.8) t²

t = 5.3 sec


consider the motion along the horizontal direction :

v₀ = initial velocity = 29 Cos62 = 13.6 m/s

a = acceleration = 0 m/s²

t = time of travel = 5.3 sec

X = horizontal displacement =?

using the equation

X = v₀ t + (0.5) a t²

X = (13.6) (5.3) + (0.5) (0) t²

X = 72.1 m


d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m

t = time taken = 5.3 sec

v = speed of center fielder

using the equation

v = d/t

v = 34.9/5.3

v = 6.6 m/s

3 0
2 years ago
Other questions:
  • A laser pulse takes 2.56 seconds to travel from Earth to the Moon and return. Use this to calculate how far away the Moon is. Ho
    14·1 answer
  • A mans walks 200 meter on 20 seconds. What is the man speed?
    9·2 answers
  • Why do all the electrons in an iron atom not settle together into the lowest energy state? because.?
    6·1 answer
  • You're driving down the highway late one night at 22 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim
    13·1 answer
  • Identify the true statement. Choose one:
    10·1 answer
  • Kriste walks once around a large circle at a constant speed.
    5·1 answer
  • Tranh slowed his skateboard as he
    11·1 answer
  • Can someone help me?
    7·1 answer
  • Aight here's the question:
    9·1 answer
  • According to the nebular theory of solar system formation, which law best explains why the solar nebula spun faster as it shrank
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!