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Oxana [17]
3 years ago
7

Please show the work and steps. The answer is (3.3 s, 15 m/s)

Physics
1 answer:
Anna [14]3 years ago
8 0

Answer:

t = 3.3 s and V = 15 m/s

Explanation:

Given that,

Distance covered by the car, d = 50 m

Initial speed, u = 5 m/s

Final speed, v = 25 m/s

Firstly we can find the acceleration of the car using third equation of motion. i.e.

v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(25)^2-(5)^2}{2(50)}\\\\a=6\ m/s^2

Let t is the time taken by the car to occur this.

t=\dfrac{v-u}{a}\\\\t=\dfrac{25-5}{6}\\\\t=3.3\ s

Let v is the average velocity of the car. It can be calculated as follows :

V=\dfrac{u+v}{2}\\\\V=\dfrac{25+5}{2}\\\\V=\dfrac{30}{2}\\\\V=15\ m/s

So, the time taken by the car to occur this is 3.3 seconds and the average velocity of the car is 15 m/s.

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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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Explanation:

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ω = 20( given )

velocity = ω r = ω l

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