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SVEN [57.7K]
3 years ago
12

The gravitational acceleration on the moon is about one-sixth the size of the gravitational acceleration on Earth. According to

Newton’s second law of motion, what happens to an astronaut who goes to the moon?
Physics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

The weight of the astronaut on the surface of moon due to the force of gravitation becomes one-sixth of the size of the gravitational acceleration on Earth.

Explanation:

Let the mass of the astronaut is, m = M

The acceleration due to gravity on Earth is, a = g

The acceleration due to gravity on the moon is a' = g/6

The weight of the astronaut on Earth, W = Mg

The weight of the astronaut on the moon is, w = Mg/6

                                                                           W = w /6

Hence, the weight of the astronaut on the surface of moon due to the force of gravitation becomes one-sixth of the size of the gravitational acceleration on Earth.

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given a circuit powered at 12V with R1, R2, R3 respectively of 10,20,30 Ohm, determine R4 in such a way that the Wheatstone brid
dalvyx [7]

Answer:

The balanced condition for Wheat stones bridge is  

Q

P

​  

=  

S

R

​  

 

as is obvious from the given values.

No, current flows through galvanometer is zero.

Now, P and R are in series, so

Resistance,R  

1

​  

=P+R

=10+15=25Ω

Similarly, Q and S are in series, so

Resistance R  

2

​  

=R+S

=20+30=50Ω

Net resistance of the network as R  

1

​  

 and R  

2

​  

 are in parallel

i=  

R

V

​  

=  

50

6×3

​  

=0.36 A.

Explanation:

8 0
3 years ago
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PolarNik [594]

OPTION C is the correct answer.

6 0
3 years ago
To determine which one of the products in a precipitation reaction is the precipitate, which of the following should you follow?
alexgriva [62]

Answer: Solubility Rules

Explanation: I took it on Ap3x

5 0
4 years ago
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Answer:

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8 0
3 years ago
A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 161.0 kmph. The cond
lukranit [14]

Answer:

384.6 m

Explanation:

The length of the runway airport should be long enough to accommodate the aircraft during its acceleration from rest to 161 km/h at rate of 2.6 m/s. We can use the following equation of motion to solve for this:

v^2 - v_0^2 = 2a\Delta s

where v0 = 0 m/s is the initial velocity of the airplane when it start accelerating, v = 161 km/h = 161*1000*(1/60)(1/60) = 44.72 m/s is the airborn speed, a = 2.6 m/s2 is the acceleration, and \Delta s is the distance of the runway, which we care looking for

44.72^2 - 0 = 2*2.6\Delta s

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4 years ago
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