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NeTakaya
2 years ago
10

How can one explain the fact that nearly all other galaxies appear to be moving farther away from us?

Physics
2 answers:
Leokris [45]2 years ago
6 0
<span>How can one explain the fact that nearly all other galaxies appear to be moving farther away from us?

</span>
<span>a) the universe is expanding

</span>
juin [17]2 years ago
4 0
A) the universe is expanding
Every galaxy is moving away from each other - not just us. And the further they are away, the faster they are moving
You might be interested in
A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp
Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \frac{sin \theta}{cos \theta} = sin θ

           sin θ = y / L

we substitute

          d \  \frac{y}{L} = m λ

with the initial data we look for the distance between the lines

           d = \frac{m \lambda \ L}{y}

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

         

we reduce nm

           λ = 396.7 nm

8 0
2 years ago
A circuit is built based on the circuit diagram shown. What is the current in the 50 Ω resistor
In-s [12.5K]

Answer:

1.2 A

Explanation:

From the diagram attached, The three resistors are parallel because the each ends of the resistors are connected together. Since they are in parallel, the voltage across each resistor is the same. The voltage source connected in parallel to the resistors is 60 V. Therefore the voltage across the 50 Ω resistor is 60 V. Using ohm law:

Voltage (V) = Current (I) × Resistance (R)

V = IR

I = V/R

I = 60 V/ 50 Ω

I = 1.2 A

The current in the 50 Ω resistor is 1.2 A

5 0
2 years ago
The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum
NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars g_{m}=3.7\ m/s^2

Time Period of pendulum on earth T=1\ s

Time period of Pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

for earth

1=2\pi\cdot \sqrt{\frac{L}{9.8}}

L=\frac{9.8}{(2\pi )^2}

L=0.498\ m

(b)For same time period on mars length is given by

L'=\frac{g_m}{(2\pi )^2}

L'=\frac{3.7}{39.48}

L'=0.0936\ m

L'=9.36\ cm                            

3 0
3 years ago
Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? explain.
choli [55]

Answer

Hi,

The forces are; weight (gravity), Normal/centripetal force and friction. Force due to gravity is constant where as friction and centripetal are not.

Explanation

Weight is constant, given by the force of gravity on the object. The centripetal force is a function of the angles occurring between the velocity vector and the weight vector that is at right angle with the perpendicular line drawn from the surface. Friction is a function of the centripetal force thus it also varies.

Hope this helps!

6 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
2 years ago
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