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MissTica
2 years ago
5

A body displaces 50cm^3 of water. What is the up-thrust on the body? ​

Physics
1 answer:
SVEN [57.7K]2 years ago
3 0

Answer:

Force of buoyancy is given by

Fb = Weight of displaced water

= mg

But, m=density x volume

= ρV g

where ρ = density of the water

V = volume of displaced water = volume of block

g = acceleration due to gravity

Given that,

V = 50 cm3

g = 981 cm/s2

ρ = 1 g/cm3

Fb = 1*50*981

= 49050 dyne

= 0.49 N

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Vector → A points due north and vector → B points due west. Both vectors have the same magnitude. Which of the following vectors
-BARSIC- [3]

Answer:

B - A

Explanation:

For the combination of 2 vector to due southwest, 1 vector must due south and the other vector due west. Since vector B is already due west, vector A should due south. As vector A is already due north, vector -A would due south. So the combination of B + (-A) or B - A should points southwest

5 0
3 years ago
A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane
kondor19780726 [428]

The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

<h3>How to find the final temperature of the gas?</h3>
  • Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
  • In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
  • The membrane is raptured without applying any external force, thus, dW=0.
  • We have the first law of thermodynamic expression as,

                                dU=dQ-dW

  • Here it is zero.

                                  dU=0,

  • As we know that,

                             dU=C_pdT=0\\\\thus,  dT=0\\\\or , T=constant\\\\i.e, T_1=T_2

  • Thus, the final temperature of the system will be equal to the initial temperature,

                          T_1=T_2=100^0C=373K

<h3>How much work is done?</h3>
  • We found that the process is isothermal,
  • Thus, the work done will be,

                               W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J

Where, R is the universal gas constant.

<h3>What is a reversible process?</h3>
  • Any process which can be made to proceed in the reverse direction is called reversible process.
  • During which, the system passes through exactly the same states as in the direct process.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

Learn more about thermodynamic processes here:

brainly.com/question/28067625

#SPJ1

7 0
2 years ago
Read 2 more answers
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

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What are the differences between concave lens and concave mirrors?
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Explanation:

A concave mirror can form real, inverted images of various sizes and virtual, erect and enlarged images whereas a concave lens forms only virtual, errect and diminished images.

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It is c convection because convection is when something is hot therefore less dense material to rise and cooler material to sink under the influence of gravity
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