All categories

3 years ago

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.

You might be interested in

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

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:

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:

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

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:

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

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

#LearnwithBrainly

0 0

3 years ago

Which of these descriptions best defines the solar system?

Marrrta [24]

Answer:

Explanation:

d because the sun is in the center and everything else surrounds it

5 0

3 years ago

Explain and derive the equation for capillary action in the phenomenon of surface tension

lapo4ka [179]

Answer:

Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.

3 0

3 years ago

Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro

ehidna [41]

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

$T=2\pi\sqrt{\frac{l}{g} }$

angular frequency ω = 2π / T

= $\omega=\sqrt{\frac{g}{l} }$

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 ( ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

4 0

3 years ago

A basic statement established by experiment or observation.

Tems11 [23]

Answer:

c.Law

Explanation:

i think you must learn and write however good luck

4 0

3 years ago

Read 2 more answers