2 years ago

What produces magnetic fields?

Physics

1 answer:

Darina [25.2K]2 years ago

6 0

B. moving electric charges, hope this helps :)

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A +0.05 C charge is placed in a uniform electric field pointing downward with a strength of 100 Newtons over Coulombs.. Determin

shtirl [24]

The magnitude of the electric force on the charge is 5 N.

<h3>Magnitude of force on the charge</h3>

The magnitude of force on the charge is calculated as follows;

F = Eq

where;

E is electric field
q is magnitude of the charge

F = 100 N/C x 0.05 C

F = 5 N

Thus, the magnitude of the electric force on the charge is 5 N.

Learn more about electric force here: brainly.com/question/20880591

#SPJ1

8 0

1 year ago

An ice skater is spinning at 6.00 rev/s with his moment of inertia being 0.400 kg/m2. Calculate his new moment of inertia if he

labwork [276]

Answer:

New moment of inertia will be $I=1.92kgm^2$

Explanation:

It is given initially angular velocity $\omega =6rev/sec=6\times 2\pi =37.68rad/sec$

Moment of inertia $I=0.4kgm^2$

Angular momentum is equal to $L=I\omega =37.68\times 0.4=15.072kgm^2/sec$

Now angular velocity is decreases to $\omega =1.25rev/sec=1.25\times 2\times 3.14=7.85rad/sec$

As we know that angular momentum is conserved

So $15.072=I\times 7.85$

$I=1.92kgm^2$

So new moment of inertia will be $I=1.92kgm^2$

4 0

3 years ago

In a physics lab experiment, a spring clamped to the table shoots a 22g ball horizontally. When the spring is compressed 21cm ,

worty [1.4K]

a square and a triangle I think

6 0

3 years ago

PLEASE HELP MEE PLEASE I BEG

TEA [102]

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

8 0

2 years ago

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob

lorasvet [3.4K]

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N$

Therefore, the force of attraction between the two charges is 3.55 N.

3 0

2 years ago