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3 years ago

Can sound travel through space? Why or why not?

Physics

1 answer:

DiKsa [7]3 years ago

5 0

Answer:

I think sound does not travel at all in space. The vacuum of outer space has essentially zero air. Because sound is just vibrating air, space has no air to vibrate and therefore no sound. If you are sitting in a space ship and another space ship explodes, you would hear nothing.

Explanation:

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A total electric charge of 6.75 nC is distributed uniformly over the surface of a metal sphere with radius 20.0 cm. If the poten

djverab [1.8K]

Answer:

a) 60 V

b) 125 V

c) 125 V

Explanation:

Given

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.

Required

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

Solution

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4 $\pi$ ∈_o (1)

r is the distance where the potential is measured and the term 1/4 $\pi$ ∈_o equals 9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m

V= 1*q/4 $\pi$ ∈_o*r

=60 V

(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by

V = 1*q/4 $\pi$ ∈_o*R (2)

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m

V = 1*q/4 $\pi$ ∈_o*R

= 125 V

(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)

V= 125 V

7 0

3 years ago

Scientific advances have led to decreased energy efficiency and waste reduction. please select the best answer from the choices

Andru [333]

It is false that scientific advances have led to decreased energy efficiency and waste reduction.

3 0

3 years ago

Read 2 more answers

Suppose a soup can is made from a sheet of steel19 which is .13 mm thick. If the can is 11 cm high and 6 cm in diameter, use dif

Flura [38]

Answer:

The can mass is 0,00359 kg or 3,59 g

Explanation:

1. Relevant Data:

Steel thickness= 0.13 mm or 0.013 mm

h=11 cm

d=6 cm

ρ=800 kg/m^3

2. Calculate mass from densisty equation:

$\rho=\frac{m}{v}$ , then $m=\rho .v$

We need to estimate the volume of the can to calculate the mass.

3. Estimate volume using differentials:

Cylinder volume equation is:

$V=\frac{1}{4}\pi d^{2}h$

Considering that the can is an object with a hole inside, then we need to estimate the real volume of the sheet of steel.

Using differentials we have:

$dV=\frac{1}{2}\pi Dh (dD)$

Then, we could say that $dD=0.013 cm$

Replacing the values of d, h and dD, we obtain:

$dV=\frac{1}{6}\pi (6 cm)(11 cm)(0,013 cm)$

$dV=0,4492 cm^3$

4. Calculate the mass

Convert volume unit into $m^3$

$0,4492 cm^3x\frac{1 m^3}{1x10^6 cm^3} =0,4492 x 10^-6 m^3$

Calculate mass

$m=\rho .v$

$m=8000 \frac{kg}{m^3}.0,4492 x10^-6 m^3$

$m=0,00359 kg =3,59 g$

5 0

4 years ago

If the mass of an object on earth is 40 kilograms it’s mass on the moon is

g100num [7]

Speaking bout mass,
it's mass is still 40Kg on the moon

but it's Weight on the moon is 1/6mg

5 0

4 years ago

A roundabout is a type of playground equipment involving a large flat metal disk that is able to spin about its center axis. A r

Zina [86]

Answer:

116.1 kgm²/s

1.12718 rad/s

Decreases

Explanation:

m = Mass of girl = 43 kg

M = Mass of roundabout = 120 kg

v = Velocity of roundabout = 2.7 m/s

r = Radius of roundabout = 1 m = R

I = Moment of inertia

Her angular momentum

$L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s$

Magnitude of angular momentum is 116.1 kgm²/s

Here the angular momentum is conserved

$L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s$

Angular speed of the roundabout is 1.12718 rad/s

Initial kinetic energy

$K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J$

Final kinetic energy

$K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J$

The overall kinetic energy decreases as can be seen. This loss is converted to heat.

5 0

3 years ago