A 2000-kg truck traveling at a speed of 6.0 m/s slows down to 4.0 m/s along a straight road. What

Mechanical (sound) waves to Earth from satellites. How is this possible? aves are unable to travel through a vacuum, such as thr

Anuta_ua [19.1K]

Sound waves are not able to travel in a vacuum, sound requires a medium such as air or a solid for example. Satellites are indeed in space, but radio waves are not sound waves they are a form of light. Light can travel in a vacuum so the messages that satellites beam back to Earth are light waves not sound waves, that is why it is possible.

6 0

3 years ago

URGENT PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST !!!

Svetlanka [38]

Answer:

An electrical current

Explanation:

An electrical current

4 0

2 years ago

A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0

3 years ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

5 0

2 years ago

Why does the surfaces of the moon, mars and earth vary in appearance

Murrr4er [49]

Answer:

due to its distance ..............

4 0

3 years ago