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tresset_1 [31]
1 year ago
11

Which sentence describes Newton's third law?

Physics
1 answer:
Elena-2011 [213]1 year ago
7 0

Option D. The image shows a pattern made by magnetic field lines around the magnet.

<h3>What is Newton's third of motion?</h3>

Newton's third of motion states that for every action, their is equal and opposite reaction.

When a magnet is placed in a magnetic field, it creates magnetic field lines because it is the region of space where its influence is felt.

Thus, the image shows a pattern made by magnetic field lines around the magnet.

Learn more about magnetic field here: brainly.com/question/7802337

#SPJ1

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Answer this plsss<br> A mobile phone operates at 900 MHz. <br> What wavelength does it use?
Aleks [24]

Answer:

The wavelength = 0.3333 meters at 900 MHz, therefore, = /4 = 0.08333 meters.

7 0
2 years ago
A 180-ohm resistor has 0.10 A of current in it. what is the potential difference across the resistor
Firlakuza [10]
We know V=IR (Ohm's law).

We are given R=180Ω and I=0.1A, then V=(0.1AΩ)(180Ω). Therefore

V=18V
5 0
3 years ago
If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa
shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

g=G\frac{m}{r^{2}}

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

g_{E}=G\frac{m}{r_{E}^{2}}

g_{P}=G\frac{m}{r_{P}^{2}}

The problem tells us the radius of the planet is twice that of the radius on earth, so:

r_{P}=2r_{E}

If we substituted that into the gravity of the planet equation we would end up with the following formula:

g_{P}=G\frac{m}{(2r_{E})^{2}}

Which yields:

g_{P}=G\frac{m}{4r_{E}^{2}}

So we can now compare the two gravities:

\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}

When simplifying the ratio we end up with:

\frac{g_{P}}{g_{E}}=\frac{1}{4}

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0
3 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
What equation explains the relation between amperes, watts and volts?
I am Lyosha [343]
I think Option B, i’m sorry if that’s wrong.
7 0
2 years ago
Read 2 more answers
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