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Marrrta [24]
3 years ago
14

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

s?
A.Wavelengths measured would match the actual wavelengths emitted.
B.Wavelengths measured would be shorter than the actual wavelengths emitted.
C.Wavelengths measured would be longer than the actual wavelengths emitted.

If a star is moving toward us in space, what would we observe about the wavelength from these stars?

A.Wavelengths measured would be shorter than the actual wavelengths emitted.

B.Wavelengths measured would match the actual wavelengths emitted.

C.Wavelengths measured would be longer than the actual wavelengths emitted.
Physics
1 answer:
torisob [31]3 years ago
7 0
There's no such thing as "stationary in space".  But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>

</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

</span>In order to decide what's actually happening, and how that star is moving, 
the trick is:  How do we know the actual wavelengths the star emitted ?


 
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A 75 kg astronaut floating in space throws a 5 kg rock at 5 m/sec. How fast does the astronaut move backwards?
JulijaS [17]

Answer:

The astronaut will get a velocity 0.064ms−1 opposite to the direction of the object.

4 0
2 years ago
What is the correct answer?
pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

    -  x^3-4x^2             <-- (x-4)(x^2)

_________________

              6x^2-16x-32

           -  6x^2-24x     <-- (x-4)(6x)

_________________

                          8x-32

                       -  8x-32 <- (x-4)(8)

___________________________

                                 0 | x^2+6x+8

This means the answer is B) x^2+6x+8

             

3 0
3 years ago
a 42.3 kg girl and a 7.93 kg sled are on the surface of a frozen lake, 15.0m apart and linked by a rope, but not moving yet. the
ycow [4]

Answer:

they meet from the girl's original position at: 2.37 (meters)

Explanation:

We need to use the Newton's law, exactly the second law that relate force, mass and acceleration as: F=m*a with this we can get both accelerations; solving for acceleration a=\frac{F}{m}. Now a_{girl}=\frac{5.76}{42.3}=0.14 (m/s^{2}) anda_{sled}=\frac{5.76}{7.93}=0.73(m/s^{2}). Then knowing that they both travel at the same time and assuming that the distance among the girl and the sled is: 15.0-x, so, x=\frac{1}{2}*a_{girl}*t^{2} and15.0-x=\frac{1}{2}*a_{sled}*t^{2}, solving for the time we get:t=\sqrt{\frac{2x}{a_{girl} } } and t=\sqrt{\frac{2*(15.0-x)}{a_{sled} } } with this equations we solving for the x that is the distance between the girl and the sled after the apply the force and we get:\sqrt{\frac{2x}{a_{girl}}} = \sqrt{\frac{2*(15.0-x)}{a_{sled} }. Finally we get:\frac{x}{a_{girl} }=\frac{(15.0-x)}{a_{sled} } and replacing the values we have got:\frac{x}{0.14} =\frac{(15.0-x)}{0.73} so 5.33*x=15-x so x=2.37 (meters).

5 0
3 years ago
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
Convert 0.0375 kg to g
leonid [27]

Answer: 37.5 g

Explanation: you multiply 0.0375 by 1000 which equals 37.5

8 0
3 years ago
Read 2 more answers
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