Which formula is used to find an object’s acceleration?

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

3 0

4 years ago

The material through which a mechanical wave travels is a.a medium.b.empty space.c.ether.d.air

Elza [17]

<h2>Answer: Medium </h2>

The medium is the main factor that differentiates a mechanical wave from an electromagnetic wave, since the first can not propagate without its existence, while the second can propagate regardless of whether the medium exists or not.

In addition, it is the medium that will define, the propagation speed of the wave, according to its specific physical characteristics.

Therefore, the <u>correct answer</u> is a.

5 0

3 years ago

When a wave is refracted, Group of answer choices it enters a new medium or material and the wave bends and changes speed. An ex

Aleonysh [2.5K]

Answer: When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.

Explanation:

6 0

3 years ago

Sergei uses a lever to lift a heavy rock. He obtains a 2.5m lever and places the fulcrum 0.7m from the rock. What is the ideal m

Anit [1.1K]

Answer:

m = 3.57

Explanation:

Given that,

Sergei uses a lever to lift a heavy rock. He obtains a 2.5m lever and places the fulcrum 0.7m from the rock.

We need to find the ideal mechanical advantage of Sergei's lever.

It is equal to the ratio of resistance arm to the effort arm. In terms of length it is given by :

$m=\dfrac{d_2}{d_1}\\\\m=\dfrac{2.5}{0.7}\\\\=3.57$

So, the ideal mechanical advantage of the lever is 3.57.

4 0

3 years ago

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance

Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

t = 5 s is the time

So, the final velocity is

$v=0+(2)(5)=10 m/s$

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

v = 10 m/s is the final velocity

Solving for s,

$s=\frac{10^2-0^2}{2(2)}=25 m$

3 0

3 years ago