Physics question? please see photo above

A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o

bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = $\omega=7.27\times 10^{-5} rad/s$

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

$v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s$

So, the velocity of the satellite is 3.08 km/s.

6 0

3 years ago

For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th

ICE Princess25 [194]

Answer:

$x=0.0498$

$x'=0.659$

Explanation:

Specific Volume $V=0.2m_3/kg$

Absolute Pressure (a) $P_a= 40kpa$

Giving

$T_a=75.87$

$v_f=1.265*10^{-3}m^3/kg$

$v_g=3.993m^3/kg$

(b) $P_a= 630kpa$

Giving

$T_b=160.13C$

$v_f'=1.10282*10^{-3} m^3/kg$

$v_g'=0.30286 m^3/kg$

(a)

Generally the equation for quality of Steam X is mathematically given by

$x=\frac{v-v_f}{v_g-v_f}$

$x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}$

$x=0.0498$

(b)

Generally the equation for quality of Steam X is mathematically given by

$x'=\frac{v-v_f'}{v_g'-v_f'}$

$x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}$

$x'=0.659$

4 0

3 years ago

Will the velocity of the book change as it moves across the surface with NO friction? Explain your answer.

MakcuM [25]

No velocity will not be changed

Why?

According to Newtons 1st law the velocity of a moving object remains unchanged unless a external force affect that.

6 0

3 years ago

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A 75-m-long train begins uniform acceleration from rest. the the train has a speed of 23 m/s when it passes a railway worker tan

Lapatulllka [165]

Answer:

acceleration a = 1.04 m/s2

Explanation:

Assume the train has a speed of 23m/s when the last car passes the railway workers. Once this happens the last car would have traveled a total distance of the 180m distance between the railway worker standing 180 m from where the front of the train started plus the 75m distance from the first car to the last car:

s = 75 + 180 = 255 m

We can use the following equation of motion to find out the distance traveled by the car:

$v^2 - v_0^2 = 2as$ where v = 23 m/s is the velocity of the car when it passes the worker, $v_0$ = 0m/s is the initial velocity of the car when it starts, a m/s2 is the acceleration, which we are looking for.

$23^2 - 0^2 = 2*a*255$

$510a = 529$

$a = 529 / 510 = 1.04 m/s^2$

8 0

4 years ago

A transverse wave is moving toward the right in a uniform medium. Point X represents a particle of the uniform medium. Which dia

trapecia [35]

After looking at the transverse waves in the diagram you listed above, the one diagram that does represent the direction of particle X at the instant show in diagram number 3. The direction of the wave motion is up. The correct answer choice will be 3.

6 0

3 years ago

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