3 years ago

What happens to the magnetic force as an object gets further away from the magnetic field?

Physics

1 answer:

sashaice [31]3 years ago

4 0

The strength of magnetic field decreases as we move away from a magnet.

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To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
$V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V$

8 0

3 years ago

As an object falls freely toward the earth, the momentum of the object-earth system (1) decreases (2) increases (3) remains the

Brums [2.3K]

The best and most correct answer among the choices provided by your question is the third choice or number 3.

<span>As an object falls freely toward the earth, the momentum of the object-earth system remains the same.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

3 years ago

When the rubber sheet is pulled down, the volume of the bell jar is ____________ and the pressure in the bell jar is ___________

Makovka662 [10]

Answer:

increases, decreases, respiration

8 0

2 years ago

A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.

Savatey [412]

Answer:

The answer should be C. slanted upward to the right.

Hope this helps. :-)

7 0

3 years ago

Read 2 more answers

A 2.0 kg block has a rope attached to the block on a table and is pulled with a force of 8.0 N. The block accelerated at 2.5m/s^

Masja [62]

Answer:

0.15

Explanation:

Assuming the rope is horizontal, sum the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum the forces in the x direction:

∑F = ma

F − Nμ = ma

Substitute:

F − mgμ = ma

mgμ = F − ma

μ = (F − ma) / (mg)

Plug in values:

μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)

μ = 0.15

3 0

3 years ago