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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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Inga [223]

Answer:

c. Momentum is the product of mass and velocity

e. Momentum is a vector quantity

g. Momentum has unit of kgm/s

Explanation:

Linear momentum P

P = m .v

m =mass

v=Velocity

If mass take in kg and velocity is in m/s then momentum p will be in kg.m/s.

1. momentum is the product of velocity and mass.

2.Momentum is a vector quantity.

3.Momentum have kg.m/s unit.

So the following option are correct.

c. Momentum is the product of mass and velocity

e. Momentum is a vector quantity

g. Momentum has unit of kgm/s.

Note-

1.Joule is the unit of energy.

2.One-half the product of mass and the square of the object's speed is known as kinetic energy.

7 0
3 years ago
The distance between two charged objects is doubled. What happens to the electrostatic force between the two?a)It will double.b)
zzz [600]

Answer:

d) It will be cut to a fourth of the original force.

Explanation:

The magnitude of the electrostatic force between the charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges of the two objects

r is the separation between the two objects

In this problem, the initial distance is doubled, so

r' = 2r

Therefore, the new electrostatic force will be

F=k\frac{q_1 q_2}{(r')^2}=k\frac{q_1 q_2}{(2r)^2}=\frac{1}{4}(k\frac{q_1 q_2}{r^2})=\frac{1}{4}F

So, the force will be cut to 1/4 of the original value.

5 0
3 years ago
Which of the following are correct statements about the way an atom is put
Ann [662]

Answer:

valenc e shell

Explanation:

4 0
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How can light energy solve our real life problem?
kifflom [539]

Answer:

It gives our light which we need for probably everything.

Explanation:

4 0
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It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w
Charra [1.4K]

Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

F = Eq

And we also know that :

F = Bqv

F = \frac{mv^{2} }{r}

Bqv = \frac{mv^{2} }{r}

or Eq = \frac{mv^{2} }{r}

Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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3 years ago
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