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ira [324]
4 years ago
6

A spacecraft is flying away from the moon toward earth. The moon's gravitational pull on the spacecraft will_______

Physics
2 answers:
Anika [276]4 years ago
4 0

the answer would be

d) decrease

tangare [24]4 years ago
4 0

Answer:

d) decrease

Explanation:

According to the law of universal gravitation, the gravitational force exerted by the moon on the spacecraft is equal to the product of their masses and inversely proportional to the square of the distance that separates them. Therefore, as the spacecraft moves away, its distance increases and the force of attraction exerted by the moon decreases.

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Which object will have the most potential energy?
Svetradugi [14.3K]

Answer:  the most potential energy ==  5 kg book, 2 m from the ground= 98 Joules

Explanation:

potential energy = m g h

m = mass

g = acceleration due gravity  = 9.8 m/s²

h = distance above  ground

1.  Pe₁ = 1 kg x 2 m x g  = 2 g

2. Pe₂ = 5 kg x 2 m x g = 10 g = 10 kg m x 9,8 m/s² = 98 Joules

3. Pe₃ = 1 kg x 0,5 m x g = 0,5 g

4. Pe₄ = 5 kg x 0.5 m x g = 2,5 g  

10 > 2,5 > 2 >0,5

5 0
3 years ago
The exosphere is the layer of the atmosphere
gayaneshka [121]
The exosphere is the layer of the atmosphere "Where gas molecules can be exchanged between Earth's atmosphere and outer space." Thus, the answer would be C.
8 0
3 years ago
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A piano emits frequencies the range from a low of about 28 Hz to a high of about 4200 Hz. Find the range of wavelengths in air a
Step2247 [10]
V=wave velocity , <span>f= frequency, </span><span>λ=wavelength </span>
<span>Use it to find corresponding wavelengths for</span><span> f=28 Hz </span>
<span>λ= v/f= 337/28=12.036 m 
</span>
<span>for f=4200 Hz </span>
<span>λ= v/f=337/4200= 0.08 m </span>
<span>So max. wavelength is 12.036 m and </span>
<span>Min Wavelength is 0.08 m </span>
<span>So the range is between .08 m and 12.036 m
</span>Hope this helps. 
4 0
4 years ago
Is a star's emission spectrum unique?
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4 0
3 years ago
The index of refraction for a certain type of glass is 1.641 for blue light and 1.603 for red light. When a beam of white light
exis [7]

Answer:

The angle between the emergent blue and red light is 0.566^{o}

Explanation:

We have according to Snell's law

n_{1}sin(\theta _{i})=n_{2}sin(\theta _{r})

Since medium from which light enter's is air thus n_{1}=1

Thus for blue incident light we have

1\times sin(40.05)=1.641\times sin(\theta _{rb})\\\\\therefore \theta _{rb}=sin^{-1}(\frac{sin(40.05)}{1.64})\\\\\theta _{rb}=23.10

Similarly using the same procedure for red light we have

1\times sin(40.05)=1.603\times sin(\theta _{rr})\\\\\therefore \theta _{rr}=sin^{-1}(\frac{sin(40.05)}{1.603})\\\\\theta _{rr}=23.66^{o}

Thus the absolute value of angle between the refracted blue and red light is

|23.66-23.10|=0.566^{o}

6 0
4 years ago
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