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DiKsa [7]
3 years ago
5

One charge is decreased to one-third of its original value,

Physics
2 answers:
Ann [662]3 years ago
3 0

Answer:

the Third option

Explanation:

Sveta_85 [38]3 years ago
3 0

Answer:

It will decrease to one-sixth the original force.

Explanation:

The electrical force is given by :

F=k\dfrac{q_1q_2}{r^2}..............(1)

Where

k is the electrostatic constant

r is the distance between charges

One charge is decreased to one-third of its original value,  and a second charge is decreased to one-half of its  original value.

i.e. q_1'=\dfrac{q_1}{3}

and q_2'=\dfrac{q_2}{2}

New force is given by :

F'=k\dfrac{q_1q_2}{6r^2}

F'=\dfrac{F}{6}

So, the new force will decrease to one-sixth the original force. Hence, the correct option is (c).

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Why do the constellations seem to move around in the sky?.
Gnoma [55]

Answer: As Earth spins on its axis, we, as Earth-bound observers, spin past this background of distant stars. As Earth spins, the stars appear to move across our night sky from east to west, for the same reason that our Sun appears to “rise” in the east and “set” in the west.

Explanation:

8 0
3 years ago
Provide the following information for each of the three types of radiations from naturally radioactive materials. Be sure to inc
scoundrel [369]

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Charge  . . . +2 elementary charges

Mass  . . . 4 Atomic Mass Units

Relative penetrating power  . . . low

The effect an electric field would have on it . . . Since the alpha particle has a positive charge, it's repelled by other positive charges, and attracted toward negative charges.

<em>Beta radiation</em>:  Particles.  Each beta particle is an electron.

Charge  . . . -1 elementary charge

Mass  . . . 0.00055 AMU

Relative penetrating power  . . . medium

The effect an electric field would have on it . . .  repelled by other negative charges, and attracted toward positive charges.

<em>Gamma radiation</em>:  electromagnetic wave, verrrrry short wave, high frequency

Charge . . . electromagnetic wave, no charge

Mass  . . . electromagnetic wave, no mass

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The effect an electric field would have on it . . . electromagnetic wave, no effect

5 0
3 years ago
The surface of the Sun has a temperature of about 5 800 K. If the radius of the Sun is 7 × 108 m, determine the power output of
Nimfa-mama [501]

Answer:

a. 3.95\times10^{26}W

Explanation:

T = temperature of the surface of sun = 5800 K

r = Radius of the Sun = 7 x 10⁸ m

A = Surface area of the Sun

Surface area of the sun is given as

A = 4\pi r^{2} \\A = 4(3.14) (7\times10^{8})^{2}\\A = 6.2\times10^{18} m^{2}

e = Emissivity = 1

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the sun is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (1) (6.2\times10^{18}) (5800)^{4}\\P = 3.95\times10^{26} W

4 0
3 years ago
Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
s2008m [1.1K]

Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

8 0
3 years ago
PLEASE HELP!!<br><br> i’ll mark brainliest if you’re correct
Leno4ka [110]
5Newtons or 5N

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2 years ago
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