An upward force applied by a fluid on an object in the fluid is the

an optician uses a plane mirror to help him. suppse a patient sits in a chair 2.5m away from him. He views the image of a chart

viva [34]

Answer:

I think 75 m

Explanation:

tell if it was correct

5 0

3 years ago

Water flowing through a garden hose of diameter 2.76 cm fills a 20.0-L bucket in 1.45 min. (a) What is the speed of the water le

mafiozo [28]

Answer:

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

Explanation:

Given;

Diameter of hose d = 2.76 cm

Volume filled V = 20.0 L = 20,000 cm^3

Time t = 1.45 min = 105 seconds

The volumetric flow rate of water is;

F = V/t = 20,000cm^3 ÷ 105 seconds

F = 190.48 cm^3/s

The volumetric flow rate is equal the cross sectional area of pipe multiply by the speed of flow.

F = Av

v = F/A

Area A = πd^2/4

Speed v = F/(πd^2/4)

v = 4F/πd^2 ......1

Substituting the given values;

v = (4×190.48)/(π×2.76^2)

v = 31.83767439628 cm/s

v = 31.84 cm/s or 0.318 m/s

the speed of the water leaving the end of the hose is 31.84 cm/s or 0.318 m/s

3 0

3 years ago

Isla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have differen

Mashcka [7]

Answer:

The answer is B) Isla had a different time than hazel

Explanation:

5 0

4 years ago

A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the

Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = $9.40\times 10^{4}\ Pa$

Pressure in the smaller pipe, P' = $2.80\times 10^{4}\ Pa$

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

$\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'$

$18^{2}v = 9^v'$

v' = 4v

Now,

By using Bernoulli's eqn:

$P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'$

where

h = h'

$\rho = 10^{3}\ kg/m^{3}$

$9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}$

$6.6\times 10^{4} = \frac{1}{2}\rho 15v^{2}$

$v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s$

Now, the rate of flow is given by:

$\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av$

$\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s$

3 0

3 years ago

Read 2 more answers

What are the optimal conditions for using a solar oven?

svetoff [14.1K]

A nice hot sunny day is what you want when using a solar oven.

5 0

4 years ago

Read 2 more answers