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Firdavs [7]
3 years ago
11

Which of these statements expresses a key assumption in our most successful models for galaxy formation?

Physics
1 answer:
Tems11 [23]3 years ago
7 0

Answer:

(B) Some regions of the universe were slightly denser than others.

Explanation:

According to the Big Bang theory, after recombination, that is, when electrons and protons first became bound to form hydrogen atoms, some slightly denser regions of the almost uniformly universe, grew gravitationally, becoming denser, forming clouds, stars and galaxies.

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Calculate the wavelength λ1 for gamma rays of frequency f1 = 7.20×1021 hz .
bagirrra123 [75]
The relationship between wavelength \lambda, frequency f and speed of light c for an electromagnetic wave is
\lambda= \frac{c}{f}
Using the data of the problem, we find
\lambda= \frac{3\cdot 10^8 m/s }{7.20 \cdot 10^{21} Hz}=4.17 \cdot 10^{-14} m
5 0
3 years ago
An astronaut having mass 320 kg with equipment included is attempting an untethered space walk. The astronaut is initially at re
ExtremeBDS [4]
This can be solved using momentum balance, since momentum is conserved, the momentum at point 1 is equal to the momentum of point 2. momentum = mass x velocity
m1v1 = m2v2
(0.03kg x 900 m/s ) = 320(v2)
v2 = 27 / 320
v2 = 0.084 m/s is the speed of the astronaut
7 0
3 years ago
How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
777dan777 [17]
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
W= qV_i - qV_f
where q is the charge of the proton, q=1 e = 1.6\cdot 10^{-19}C, with e being the elementary charge, and V_i = +125 V and V_f = -55 V are the initial and final voltage.

Substituting, we get (in electronvolts):
W=e(125 V-(-55 V))=180 eV
and in Joule:
W=(1.6 \cdot 10^{-19})(125 V-(-55V))=2.88 \cdot 10^{-17}J

5 0
3 years ago
The period of the earth around the sun is 1 year and its distance is 150 million km from the sun. An asteroid in a circular orbi
SOVA2 [1]

Answer:

5.024 years

Explanation:

T1 = 1 year

r1 = 150 million km

r2 = 440 million km

let the period of asteroid orbit is T2.

Use Kepler's third law

T² ∝ r³

So,

\left ( \frac{T_{2}}{T_{1}} \right )^2=\left ( \frac{r_{2}}{r_{1}} \right )^3

\left ( \frac{T_{2}}{1} \right )^2=\left ( \frac{440}{150} \right )^3

T2 = 5.024 years

Thus, the period of the asteroid's orbit is 5.024 years.

4 0
2 years ago
A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o
lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

K_{1} + W_{F} = K_{2}

Where:

W_{F} - Work done by the external force, measured in joules.

K_{1}, K_{2} - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

W_{F} = K_{2} - K_{1}

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})

Where:

m - Mass of the object, measured in kilograms.

v_{1}, v_{2} - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}

v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}

v_{1} \approx 25.060\,\frac{m}{s}

Final velocity

v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}

v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}

v_{2} \approx 33.121\,\frac{m}{s}

Finally, if m = 3.5\,kg, v_{1} \approx 25.060\,\frac{m}{s} and v_{2} \approx 33.121\,\frac{m}{s}, then the work done by the force is:

W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]

W_{F} = 820.745\,J

The work done by the force is 820.745 joules.

6 0
2 years ago
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