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mash [69]
3 years ago
6

Which statement does Einstein's theory promote about space and time?

Physics
2 answers:
Ganezh [65]3 years ago
6 0

Answer: Time and space depend on reference frames, but the speed of light does not.

Explanation: According to Einstein Space and Time are directly related as the way we sense them, in his words space and time is like a fabric in which we cannot affect one without affecting the other as we interact with it, time can be stretched and squeezed depending on your inertial frame of reference (constant speed and mass). However light can travel thru the fabric of space and time and it does not get affected by any reference keeping always the same speed.

Cloud [144]3 years ago
4 0

Answer:

2nd statement would be the right answer

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Water moves through a constricted pipe in steady, ideal flow. At the
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A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

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