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3 years ago

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A purse at radius 2.0 m and a wallet at radius 2.3 m travel in uniform circular motion on the floor of a merry-go-round as the r

ide turns the magnitude of the acceleration of the purse is 2.0 m/s/s. Determine the acceleration of the wallet.

1 answer:

horrorfan [7]3 years ago

4 0

Answer:

Explanation:

Given

radius of purse from center $r_p=2\ m$

radius of wallet from center $r_w=2.3\ m$

Merry - go - round is undergoing uniform circular motion therefore angular velocity is constant

acceleration of purse is equal to centripetal acceleration $a_p=\omega ^2\times r$

where $\omega =angular\ velocity$

$2=\omega ^2\times 2$

$\omega =1\ rad/s$

acceleration of wallet is

$a_w=\omega ^2\times r_w$

$a_w=1\times 2.3=2.3\ m/s^2$

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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:

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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The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

Sindrei [870]

Answer with Explanation:

We are given that

Speed , $v_0$ =107257 kmph

Angular velocity, $\omega=7.292\times 10^{-5} rad/s$

Radius of earth,r= $6371 km=6371000 m$

1 km=1000 m

Linear velocity, $v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s$

Linear velocity, $v=464.57\times \frac{18}{5}=1672.46 km/h$

Velocity at point A, $v_A=-vi+v_0 j=-1672.46 i+107257j kmph$

Velocity at point B, $v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph$

Velocity at point C, $v_C=v_0j+vi=1672.46 i+107257j kmph$

Velocity at point D, $v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph$

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3 years ago

“Is it correct to say that a radio wave can be considered a low-frequency light wave?

klio [65]

It's weird but technically correct to say that a radio wave can be considered a low-frequency light wave. Radio and light are both electromagnetic waves. The only difference is that radio waves have much much much longer wavelengths, and much much much lower frequencies, than light waves have. But they're both the same physical phenomenon.

However, a radio wave CAN'T also be considered to be a sound wave. These two things are as different as two waves can be.

-- Radio is an electromagnetic wave. Sound is a mechanical wave.

-- Radio waves travel more than 800 thousand times faster than sound waves do.

-- Radio waves are transverse waves. Sound waves are longitudinal waves.

-- Radio waves can travel through empty space. Sound waves need material stuff to travel through.

-- Radio waves can be detected by radio, TV, and microwave receivers. Sound waves can't.

-- Sound waves can be detected by our ears. Radio waves can't.

-- Sound waves can be generated by talking, or by hitting a frying pan with a spoon. Radio waves can't.

-- Radio waves can be generated by an alternating current flowing through an isolated wire. Sound waves can't.

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4 years ago

Bob and Lily are riding on a typical carousel. Bob rides on a horse near the outer edge of the circular platform, and Lily rides

alukav5142 [94]

Answer:

Bob's angular speed is the same as that of lily

Explanation:

Because for a carousel the angular speed remains the same since velocity at center and edge are the same

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3 years ago

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K

final pressure can be calculate by using below relation

$\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}$

we know that

force = pressure * area

therefore force is

$F =(\frac{T_{1}}{T_{2}}*P_{1})A$

$F =(\frac{378}{278}*101325)(17*10^{-2})^{2}$

F = 3.98 kN

5 0

3 years ago