PHYSICS HIGH SCHOOL

The moon's mass is 7.35 × 1022 kg, and it moves around the earth approximately in a circle or radius 3.82 × 105 km. The time required for one revolution is 27.3 days. Calculate the centripetal force that must act on the moon. How does this compare to the gravitational force that the earth exerts on the moon at that same distance?

Answers

Answer 1

Answer:

Explanation:

It is given that,

Mass of moon, $m=7.35* 10^(22)\ kg$

Radius of circle, $r=3.82* 10^(5)\ km=3.82* 10^(8)\ m$

The time required for one revolution is 27.3 days, t = 27.3 days

1 day = 86400 seconds

27.3 days = 2358720 seconds

Let v is the speed of moon around the circular path. It is given by :

$v=(2\pi r)/(T)$

$v=(2\pi * 3.82* 10^(8)\ m)/(2358720\ s)$

v = 1017.57 m/s

Let F is the centripetal force acting on the moon. It is given by :

$F=(mv^2)/(r)$

$F=(7.35* 10^(22)\ kg* (1017.57\ m)^2)/(3.82* 10^(8)\ m)$

$F=1.99* 10^(20)\ m/s^2$

So, the centripetal force that must act on the moon is $1.99* 10^(20)\ m/s^2$ . The gravitational force that the earth exerts on the moon at that same distance is also equal to $1.99* 10^(20)\ m/s^2$ . Hence, this is the required solution.

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Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of their electrical charges q1 and q2 and inversely as the square of the distance d between the particles. Find the effect on F of doubling q1 and q2 and halving the distance between them.
Examples of the period, frequency, speed of a wave in a sentence?

What is the difference in the charges on a ballon rubbed in your hair and a glass rod rubbed

Answers

The balloon steals electrons from your hair, leaving the hair positively charged, and the balloon negatively charged. It causes hair to be apart from each other, because it has the same charge. Glass has a weaker hold on electrons, and silk absorbs the lost electrons.

The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa and 30°C enters it with a velocity of 357 m/s and the exit state is 200 kPa and 90°C. The specific heat of air at the average temperature of 60°C = 333 K is cp = 1.007 kJ/kg·K.

Answers

Answer:

81.29 m/s

Explanation:

Given:

60°C = 333 K is cp = 1.007 kJ/kg·K.

at air inlet:

Pressure $P_(1)$ = 100 kPa

Temperature $T_(1)$ = 30°C

Velocity $V_(1)$ = 357 m/s

at air outlet:

Pressure $P_(2)$ = 200kPa

Temperature $T_(2)$ = 90°C

We can define the energy balance as:

$E _(in)$ - $E_(out)$ = Δ $E_(system)$ = 0

therefore, $E _(in)$ = $E_(out)$

m $h_(1)$ + m $(V_(1)^(2) )/(2)$ =m $h_(2)$ + $(V_(2)^(2) )/(2)$ (taking m common and cancelling it)

4 $h_(1)$ + $(V_(1)^(2) )/(2)$ = $h_(2)$ + $(V_(2)^(2) )/(2)$

The velocity at the exit of a diffuser:

$V_(2)$ = [ $V_(1)^(2)$ + 2 $c_p{}$ ( $T_(1)$ - $T_(2)$ ) $]^(0.5)$

$V_(2)$ = [ $357^(2)$ + 2 x 1000 x 1.007 x (30-90) $]^(0.5)$

$V_(2)$ = 81.29 m/s

therefore, the velocity at the exit of a diffuser is 81.29 m/s

Answer:

Exit velocity = 498.3m/s

Explanation:

The energy balance for the system is given by:

Ein - Eout = ◇Esystem

Ein = Eout

Therefore

m' (h1 + V1^2/2) = m' (h2 + V2^2/2)

h1 + V1^2/2 = h2 + V2^2/2

The exit velocity will be:

V2 = [V1^2 + 2(h1 - h2)^0.5 = [V1^2 + 2(cp(T1 - T2)]^0.5

But cp = 1.007kJ/kg

V2 = [(357)^2 + 2(1.007)(90-30)(1000m^2/s^2)]^0.5

V2 = [127449 + 120840]^0.5

V2 = [248289]^0.5

V2 = 498.3m/s

A thin conducting plate 50.0 cm on a side lies in the xy plane. If a total charge of 4.00 ? 10 ?8 C is placed on the plate, find (a) the charge density on the plate, (b) the electric field just above the plate, and (c) the electric field just below the plate.

Answers

Answer:

a) 8*10^-8C/m²

b) +9.04*10^3N/C

c) = -9.04*10^3N/C

Explanation:

Given

Side length, L = 50cm = 0.5m

Charge on the plate, Q = 4*10^-8C

Surface charge density, σ = Q/A

The surface charge density of each part is then half of the total charge density of the plate. Thus,

σ(face) = 1/2σ

σ(face) = Q/2A

σ(face) = Q/2L²

Now we plug in, since we have Q and L

σ(face) = 4*10^-8 / 2*0.5²

σ(face) = 4*10^-8 / 0.5

σ(face) = 8*10^-8C/m²

Magnitude of electric field above the plate is,

E = σ(face) / E•

E = 8*10^-8 / 8.85*10^-12

E = 9.04*10^3 N/C

If we assume this plate lies on the side of the "xy" plane, the electric field is directed in the positive "z" direction. As such,

E = +9.04*10^3N/C

Electric field below the plate has the same magnitude, but different direction. So, E = -9.04*10^3N/C

While in Canada, the engineer buys a new electric toothbrush. When she returns home she designs anothertransformer so she can use the toothbrush in India. This transformer also has an input coil with 240 turns.
How many turns should the output coil have?

Answers

I think 1000 but I don't really know

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Answers

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Answers

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