All categories

3 years ago

Electromagnetic force is present when electromagnetic fields

Physics

2 answers:

attashe74 [19]3 years ago

8 0

Answer:

c- interact with charged particles.

Explanation:

When a charged particle is placed in the field of the other charged particle then it will experience electromagnetic force on it.

As we know that force on a charged particle due to some other electric field is given by

F = qE

here we know

q = charge

E = electric field intensity

Similarly when a charged particle is placed in external magnetic field then the force on that moving charge is given by

$F = q(v\times B)$

so here if charge is moving in the magnetic field due to some other system then the force on it is given by above equation.

So here we can say that electromagnetic force is present when electromagnetic fields

c- interact with charged particles.

AlekseyPX3 years ago

4 0

b is the correct answer

You might be interested in

In an emergency situation, firemen need to respond as quickly as possible. If a fireman is responding from the second floor, how

rjkz [21]

Answer:

4m/s

Explanation:

May be different considering how long the pole is and how heavy the firefighter is.

8 0

3 years ago

What is the magnetic force (in newtons) on a particle traveling in a 1.5 T magnetic field if q = 7.5 microcoulombs and v = 1.75

Alla [95]

Given:

B(Magnetic field): 1.5 T

q= 7.5 microcoulombs

v= 1.75 x 10 ∧6 m/s

The angle ∅ between B and v is 45 °.

Now we know that F= qvB sin ∅

Substituting these values we get:

F= 7.5 x 10∧-6 x 1.75 x 10∧6 x 1.5 x sin 45

F= 16.752 N

3 0

3 years ago

A 6.5×10−2 kg arrow hits the target at 25 m/s and penetrates 3.8 cm before stopping.

White raven [17]

1) We use:
2as = v² - u², with v = 0,
to find the acceleration of the arrow.
2 x 0.038 x a = -(25)²
a = -8.22 x 10³ m/s²
F = ma
F = 6.5 x 10⁻² x -8.22 x 10³
F = -534.2 N; the negative direction indicates that the force is in the opposite direction of the motion.

B) The arrows force is the same but in the opposite directioin.
534.2 N

C) a = -8.22 x 10³ m/s²
s = -(65)²/(2 x -8.22 x 10³)
s = 25.7 cm

8 0

3 years ago

Suppose a 200-mm focal length telephoto lens is being used to photograph mountains 11.5 km away. Part (a) What is image distance

Nuetrik [128]

This is your answer...Down below

5 0

3 years ago

Calculate the upthrust aciting on a body if its

Zepler [3.9K]

Answer:

As a body moving upward

T=real weight + apparent weight

T=550+490

T=1040

hope u will get the answer:)

Explanation:

7 0

3 years ago