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
NeX [460]
2 years ago
14

A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram

. a field of circles labeled b. there is a positively charged particle with a velocity vector pointing 37 degrees north of east labeled v. what is the magnitude of the magnetic force acting on the charge? 6.6 × 10–3 n 4.9 × 10–3 n 4.9 × 103 n 6.6 × 103 n
Physics
1 answer:
Sauron [17]2 years ago
3 0

The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

\rm F=qvBsin\theta

where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

F=6.6\times 10^{-3}\ N

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

To know more about Magnetic force follow

brainly.com/question/14411049

You might be interested in
Which dog has the most kinetic energy
Blababa [14]
The formula to find the kinetic energy is:

Ek= 1/2 × m × v^2

1. Ek= 1/2×15×3^2
= 67.5 J

2.Ek= 1/2×8×4^2
=64 J

3.Ek= 1/2×12×5^2
= 150 J

4.Ek= 1/2×10×6^2
= 180 J

So the fourth dog has the most kinetic energy.
4 0
3 years ago
If the energy carried by a wave increases, which other wave property also increases?
damaskus [11]

Answer:

The answer is wavelength

5 0
3 years ago
Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A
Vera_Pavlovna [14]

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

7 0
3 years ago
What should i do if i'm stuck at home? Any thing except watching movies and going to sleep
Makovka662 [10]

card games

board games

bird watch

write a song

make a game

count stuff

throw a ball with someone

play outside

3 0
3 years ago
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
3 years ago
Other questions:
  • Leap years _____. happen because the Earth revolves around the sun in less than 365 days make up for the extra one-fourth day th
    15·1 answer
  • A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
    5·1 answer
  • What do neutrons and protons have in common? How are they different?
    9·2 answers
  • If an alveolus with an initial volume of 3 ml of air with a total pressure of 810 mmhg decreases in volume to 1.7 ml, what would
    13·2 answers
  • On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som
    6·1 answer
  • When kicking a football, the kicker rotates his leg about the hip joint. If the velocity of the tip of the kicker’s shoe is 35.0
    11·1 answer
  • Falling Faster
    15·1 answer
  • Light passing through a double slit
    11·2 answers
  • Someone please help me!! See PDF to help
    13·2 answers
  • Consider the following arrangement with a frictionless/massless pulley. Determine the force F required to move block A if the co
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!