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3 years ago

Which law applies to the picture?Newtons 1st Law?Newton's 2nd Law?Newton's 3rd Law?

Physics

2 answers:

Diano4ka-milaya [45]3 years ago

8 0

<h2>Answer:</h2>

It applies Newton's 1st Law.

<h3>Newton's 1st Law:</h3>

Every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

jeka57 [31]3 years ago

7 0

Answer:

Newtons 1st Law.

Explanation:

Because it states as every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

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A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

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2 years ago

As light shines from air to water, the index of refraction is 1.02 and the angle of incidence is 38.0 Â°. What is the light's an

muminat

Answer:

Light's angle of refraction = 37.1° (Approx.)

Explanation:

Given:

Index of refraction = 1.02

Base of refraction = 1

Angle of incidence = 38°

Find:

Light's angle of refraction

Computation:

Using Snell's law;

Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction

Sin38 / Light's angle of refraction = 1.02 / 1

Sin[Light's angle of refraction] = Sin 38 / 1.02

Sin[Light's angle of refraction] = [0.6156] / 1.02

Sin[Light's angle of refraction] = 0.6035

Light's angle of refraction = 37.1° (Approx.)

5 0

3 years ago

A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l

DENIUS [597]

Answer:

$\mathbf{\beta = 123.75 \ dB}$

Explanation:

From the question, using the expression:

$125 \ dB = 10 \ log (\dfrac{I}{I_o})$

where;

$I_o = 10^{-12} \ W/m^2$

$I = 10^{12.5} \times 10^{-12} \ W/m^2$

$I = 3.162 \ W/m^2$

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = $\dfrac{3.162\times 3}{4}$

= 2.372 W/m²

Thus;

$\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2$

$\mathbf{\beta = 123.75 \ dB}$

7 0

3 years ago

The temperature of a room is 77°F what would be the temperature in Celsius scale

anygoal [31]

Formulas change from F to degree C : C = ( F - 32 )/1.8

so we have (77-32)/1.8 = 25 oC

ok done. Thank to me :>

5 0

3 years ago

The light from the sun has higher frequencies from one side of the sun than from the other side. What does that tell you about t

morpeh [17]

If the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.

Doppler effect states that, if a person is standing still and a source ( sound / light ) is moving towards him, the frequency of the wave emitted from the object will increase and if the source ( sound / light ) is away from him, the frequency of the wave emitted from the object will decrease.

So, if the light from the sun has higher frequencies from one side of the sun than from the other side, it means that the Sun is rotating. The higher frequencies points are the points that rotating towards Earth and lower frequencies points are the points that rotating away from Earth.

Therefore, if the light from the sun has higher frequencies from one side of the sun than from the other side, it is proof that the Sun is rotating.

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8 0

1 year ago

Which law applies to the picture?Newtons 1st Law?Newton's 2nd Law?Newton's 3rd Law?​

Which law applies to the picture?Newtons 1st Law?Newton's 2nd Law?Newton's 3rd Law?