Ask question

Ask question

All categories

4 years ago

6

Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe

n the container is full, what is the pressure that the water exerts on the sides of the container at the bottom of the container above the atmospheric pressure? b. When the container is full, what is the pressure that the water exerts on the sides of the container halfway down from the top above the atmospheric pressure?

1 answer:

uranmaximum [27]4 years ago

7 0

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = $\pi r^2h$

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x $0.13^2$ x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

You might be interested in

If an object's mass increases, its--

aev [14]

Answer:

both kinetic and potential energy

Explanation:

this is your ans

I hope it helps mate

I will always help you understanding your assingments

have a great day

#Captainpower :)

8 0

3 years ago

Hi,please i need help with this one in physics.hurry and correct i need it.<br>Topic:Equilibrium.

Rudik [331]

The half-meter rule (easy math) is 0.5 meters or 50 centimeters since a meter is 1 meters long, which is equivalent to 100 centimeters. Therefore, we shall apply the 50 cm rule.

A 50 cm rule's center of mass is now 25 cm away.

Additionally, according to the data, the object is pivoted at 15 cm, while the 40 g object is hung at 2 cm from the rule's beginning. Using a straightforward formula, we can compare the two situations: the distance from the pivot to the center of the mass times the mass of the 40 g object divided by 2 cm must equal the distance from the pivot to the center of the mass times mass of the 10 x g object

The result of the straightforward computation must be 52g.

Most simplified version:

the center of mass of the rule is at the 25 cm mark

⇒ $40 g * (15 cm - 2 cm)$

⇒ $= M * (25 cm - 15 cm)$

#SPJ2

5 0

2 years ago

Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field

Scilla [17]

Answer:

Electric field lines point away from positive charges and toward negative charges. <u>True</u>

Electric field lines are continuous; they do not have a beginning or an ending.<u> False</u>

Electric field lines can never intersect. <u>True</u>

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

Explanation:

Electric field lines point away from positive charges and toward negative charges. Always the field lines go to negative charges and leave from positive charges.

Electric field lines are continuous; they do not have a beginning or an ending.<u> False </u>

Because the field lines starts at positive charges and ends in negative charges.

Electric field lines can never intersect. <u>True</u>

It can not intercept the field lines because in that point the the field would have two directions. Besides, in that point the real value of the field should be found adding both field lines.

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

This fact is opposite to that so the regions of space where the magnitude the electric field is weak the lines are father apart and where the field is strong the lines are close together.

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

This statement correspond to the definition of the field line.

5 0

3 years ago

A 1000-n weight is hanging from a 2.0 m long aluminum rod. A 500-n weight is hanging from a 1.0 m long aluminum rod. The two alu

Hunter-Best [27]

Answer:

Both rod have the same tensile stress

Explanation:

Given information,

The weight first rod, $W_{1}$ = 1000 N

The length of first rod, $l_{1}$ = 2.0 m

The weight second rod, $W_{2}$ = 500 N

The length of second rod, $l_{2$ = 1.0 m

The equation of tensile stress, σ = $\frac{F}{A}$

where

σ = tensile stress (N/ $m^{2}$ or Pa)

F = Force (N)

A = Area (N/ $m^{2}$ or Pa)

so

σ1 = $\frac{W_{1} }{A_{1} }$ , A = 2πl

= $\frac{1000}{2\pi(2) }$

= $\frac{250}{\pi }$ N/ $m^{2}$

now calculate σ2

σ2 = $\frac{W_{2} }{A_{2} }$

= $\frac{500}{2\pi(1) }$

= $\frac{250}{\pi }$ N/ $m^{2}$

σ1/σ2 = $\frac{250}{\pi }$ / $\frac{250}{\pi }$

σ1/σ2 = 1

σ1 = σ2

Hence, the tensile stress of first and second rod are the same.

5 0

4 years ago

In 1998, scientists discovered that the expansion of the universe has been accelerating.

Lynna [10]

Dang that’s a cool fact bro

8 0

3 years ago

Read 2 more answers