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2 years ago

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Which of the following is supportive evidence of the big bang theory? pulsars the different shapes of galaxies cosmic microwave

background radiation galactic clusters

2 answers:

Juli2301 [7.4K]2 years ago

6 0

Cosmic microwave background radiation

Lunna [17]2 years ago

3 0

Cosmic microwave background radiation

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

Compare and contrast series and parallel circuits?

dalvyx [7]

In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.

Hope It Helps!

6 0

2 years ago

Two factor that affect gravitational force

Gnoma [55]

When dealing with the force of gravity between two objects, there are only two things that are important – mass, and distance. The force of gravity depends directly upon the masses of the two objects, and inversely on the square of the distance between them.

6 0

3 years ago

Mariulka [41]

Answer:

The new volume of the balloon is 8.8 L

Explanation:

Charles's Law consists of the relationship between the volume and the temperature of a certain amount of ideal gas, which is kept at a constant pressure. This law establishes that the volume is directly proportional to the temperature of the gas, that is to say that if the temperature increases, the volume of the gas increases and that on the contrary if the temperature of the gas decreases, the volume decreases.

In other words, Charles's law states that when the amount of gas and pressure are kept constant, the ratio between the volume and the temperature will always have the same value:

$\frac{V}{T}=k$

Having a certain volume of gas V1 that is at a temperature T1, by varying the volume of gas to a new value V2, then the temperature will change to T2, and it will be fulfilled:

$\frac{V1}{T1}=\frac{V2}{T2}$

In this case:

V1= 8.5 L
T1= 294 K
V2= ?
T2= 305 K

Replacing:

$\frac{8.5 L}{294 K}=\frac{V2}{305 K}$

Solving:

$V2=305 K*\frac{8.5 L}{294 K}$

V2=8.8 L

<u><em>The new volume of the balloon is 8.8 L</em></u>

7 0

3 years ago

Consider two identical objects released from rest high above the surface of Earth. (Neglect air resistance for this question.)

Nadusha1986 [10]

The question is missing its alternatives. Here is the complete question.

Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.

1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.

a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;

Answer: C) K2 = (4/3)K1

Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - $\frac{G.Me.m}{r}$ + U₀

In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.

The potencial energy of an object of mass m on the surface of the Earth is:

Usurface = - $\frac{G.Me.m}{Re}$

The potencial energy of the object in Case 1 is

U1 = - $\frac{G.Me.m}{2Re}$

For the Case 2:

U2 = - $\frac{G.Me.m}{3Re}$

The potencial change in Case 1:

ΔU1 = - G.Me.m.( $\frac{1}{Re}-\frac{1}{2Re}$ ) = - $\frac{1}{2}$ $\frac{G.Me.m}{Re}$

For Case 2:

ΔU2 = - G.Me.m( $\frac{1}{Re}-\frac{1}{3Re}$ ) = - $\frac{2}{3}$ $\frac{G.Me.m}{Re}$

Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:

Δ $\frac{U2}{U1}$ = (-2/3)(-1/2) = 4/3

So, comparing kinetic energies, K2 is 4/3 of K1.

5 0

3 years ago