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larisa86 [58]
2 years ago
11

Calculate the acceleration due to gravity on the Moon. The Moon’s radius is 1.74 x 106 m and its mass is 7.35 x 1022 kg.

Physics
1 answer:
GarryVolchara [31]2 years ago
5 0
G=?
m=7.35*10^22kg
r=1.74*10^6m
G=6.67*10^-11nm^2kg^-2
you can use the formula
g=Gm/r^2
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A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner
sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0
2 years ago
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t
Mashutka [201]
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
f = (\frac{g + v_{r}}{g + v_{s}})f_{o} -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
v_{r} = Velocity of the receiver/observer relative to the medium = ?.
v_{s} = Velocity of the source with respect to medium = 0 m/s.
f_{o} =  Frequency emitted from source = 400 Hz.
f = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

408 = ( \frac{340 + v_{r}}{340 + 0})*400

\frac{408}{400} =  \frac{340 + v_{r}}{340}

Solving above would give you,
v_{r} = 6.8 m/s

The correct answer = 6.8m/s



7 0
2 years ago
If the magnitude of charges on these source charges is arranged in a descending order, which is the correct sequence?
bonufazy [111]
The source charges' magnitude is signified by the arrows pointing outward. The more arrows there are, the greater is its magnitude. This is because, each arrow represents an electrical force exerted by the source. When you add up all the arrows there is, the electrical force becomes even greater. The answer in descending order would be C > A > B > D.
6 0
2 years ago
Read 2 more answers
If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the fr
Dmitriy789 [7]

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance C=8.5nF=8.5\times 10^{-9}f

Potential difference across capacitor V = 12 volt

Current through capacitor i=3.33mA=3.33\times 10^{-3}A

Capacitive reactance will be equal to X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm

Capacitive reactance is equal to X_c=\frac{1}{\omega C}

3603.60=\frac{1}{\omega\times  8.5\times 10^{-9}}

\omega =32647.091rad/sec

2\pi f=32647.091

f=5198.98Hz

f = 5.20 kHz

So frequency will be equal to 5.20 kHz

So option (c) will be correct answer

3 0
2 years ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

We know that

P_1 = \frac{nRT_1}{v}

P_2 = \frac{nrT_2}{v}

n and v remain same at both side. so we have

\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}

P_1 = \frac{21}{11} P_2 ..............1

let final pressure is P and temp  T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3} ..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3} .............3

divide 2 equation by 3rd equation

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, temperature on left side is 1.48 times the temperature on right

6 0
2 years ago
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