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Igoryamba
3 years ago
6

A tank 20 m deep and 7m wide is layered with 8m of oil,6m of water and 5m of mercury.complete total hydroatatic force.(density o

f oil and mercury is 800 and 13600kg/m respectively ).​
Physics
1 answer:
professor190 [17]3 years ago
8 0

Answer:

F = 3.03 10⁷ N

Explanation:

We will eat by calculating the pressure in the tank

         P = ρ g h

the pressure totals the sum of the pressure of each liquid

       P_total = P_oil + P_water + P_Hg

       P_total = ρ_oil ​​g h_oil + ρ_water g h_water + ρ_Hg g h_Hg

       P_total = g (ρ_oil ​​h_oil + ρ_water h_water + ρ_Hg h_Hg)

       P_total = 9.8 (800 8 + 1000 6 + 13 600 5)

        P_total = 7,879 10⁵ Pa

The definition of Pressure is

       P = F / A

        F = P A

The area of ​​a tank is the area of ​​a circle

       A = π r² = π d² / 4

       

      F = P π d² / 4

let's calculate

      F = 7,879 10⁵ π 7²/4

      F = 3.03 10⁷ N

In this calculation the atmospheric pressure was not taken into account because they ask the hydrostatic pressure

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Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What
yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

<u>Step 1: </u>Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water  =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

<u>Step 2:</u> formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

<u>Step 3: </u>Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

<u>Step 4: </u>Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

<u>Step 5:</u> Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

<u>Step 6</u>: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

<u>Step 7:</u> Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0
3 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Leya [2.2K]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

6 0
3 years ago
A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente
inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

F=\frac{mv^2}{r}

Where, m is the mass of the ball, v is the speed and r is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

20.4 = \frac{3.1\times v^2}{2}

and, v^2 = 13.16

Thus, v = 3.63 m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = \frac{\text {Path length} }{\text {Speed}} = \frac{12.56}{3.63} = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0
3 years ago
Remember KE=M*V2/2…
 A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy
pishuonlain [190]

Answer: 4 times as much

Explanation:

4 0
2 years ago
(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

4 0
4 years ago
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