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Usimov [2.4K]
3 years ago
13

The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det

ermine (a) the orbital speed of the earth and (b) the mass of the sun.
Physics
1 answer:
seraphim [82]3 years ago
8 0
(a) Determine the circumference of the Earth through the equation,
            C = 2πr
Substituting the known values, 
           C = 2π(1.50 x 10¹¹ m)
             C = 9.424 x 10¹¹ m

Then, divide the answer by time which is given to a year which is equal to 31536000 s. 
          orbital speed = (9.424 x 10¹¹ m)/31536000 s

               orbital speed = 29883.307 m/s

Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg. 
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A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to de
Lorico [155]

Answer:

The correct answer is a

Explanation:

The speed of a sound wave depends on the square root of the modulus of compressibility and the density of the medium.

For the same medium, the speed of sound depends on the temperature of the fora

           v = v_o \ \sqrt{1 + \frac{T}{273} }

Therefore, the different results that are obtained are due to changes in temperature. The correct answer is a

since this way it has the values ​​of the speed of sound for each temperature, for which it can compare with the results obtained from the trip.

3 0
3 years ago
A(n) 930 N crate is being pushed across a level floor by a force of 400 N at an angle of 20◦ above the horizontal. The coefficie
Nana76 [90]

Answer:

The magnitude of the acceleration of the box is 2.291\,\frac{m}{s^{2}}.

Explanation:

The free body diagram of the crate is included as attachment, whose equations of equilibrium are described below:

\Sigma F_{x} = P\cdot \cos 20^{\circ} - \mu_{k}\cdot N = \left(\frac{W}{g}\right)\cdot a

\Sigma F_{y} = P\cdot \sin 20^{\circ} + N - W = 0

From second equation of equilibrium we find an expression for the normal force and find the respective value:

N = W - P \cdot \sin 20^{\circ}

N = 930\,N - 400\cdot \sin 20^{\circ}\,N

N = 793.192\,N

Lastly, the acceleration experimented by the crate during pushing is cleared in the first equation of equilibrium and consequently calculated:

a = \frac{P\cdot \cos 20^{\circ}-\mu_{k}\cdot N}{\frac{W}{g} }

a = \frac{400\cdot \cos 20^{\circ}\,N-0.20\cdot (793.192\,N)}{\frac{930\,N}{9.807\,\frac{m}{s^{2}} } }

a = 2.291\,\frac{m}{s^{2}}

The magnitude of the acceleration of the box is 2.291\,\frac{m}{s^{2}}.

3 0
3 years ago
What are the issues that hinders efforts to achieve sustainability?
Nadya [2.5K]

Answer:

who will solve environmental problems, who is responsible for environmental problems, and who pays the cost of implementing solutions

Explanation:

4 0
3 years ago
What could we call the<br> grocery store?<br> A. Linear motion<br> B. Reference point<br> C. Rotary
Roman55 [17]

Reference point

Explanation:

I am not sure

8 0
3 years ago
For the circuit shown in the figure(figure 1) find the current through each resistor. Express your answers using two significant
Angelina_Jolie [31]

The current flowing in each resistor of the circuit is 4 A.

<h3>Equivalent resistance of the series resistors</h3>

The equivalent resistance of the series circuit is calculated as follows;

6 Ω and 4 Ω are in series = 10 Ω

5 Ω and 10Ω are in series = 15 Ω

<h3>Effective resistance of the circuit</h3>

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\R = \frac{R_1R_2}{R_1 + R_2} \\\\R = \frac{10 \times 15}{10 + 15} \\\\R = 6 \ ohms

<h3>Current flowing in the circuit</h3>

V = IR

I = V/R

I = 24/6

I = 4 A

Learn more about resistors in parallel here: brainly.com/question/15121871

8 0
2 years ago
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