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3 years ago

What is the volume of a cone with a height of 27 cm

Physics

1 answer:

JulijaS [17]3 years ago

4 0

Explanation:

→ Volume of cone = πr² × h/3

Here,

Radius (r) = 13 cm
Height (h) = 27 cm

→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

→ Volume of cone = 33462/7 cm³

→ Volume of cone = 4780.28 cm³

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The engine oil at 150 degree Celsius is cooled to 80 degree Celsius in a parallel flow heat exchanger by water entering at 25 de

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Answer:

Explanation:

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3 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

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The rule of inertia states that an object in motion will stay in motion unless another force has acted upon it. Because the person doesn't have their seatbelt on, they will keep moving. But if they were wearing a seatbelt, that would work as the force that is supposed to stop the person from flying forward.

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3 years ago

efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.

PolarNik [594]

Answer:

Inlet : $v_i=0.0646\frac{m}{s}$

Outlet: $v_o=0.171\frac{m}{s}$

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

$\dot{m}=\frac{\dot{V}}{\upsilon}$ , where $\dot{V}$ represent the flow rate and $\upsilon$ the specific volume at the pressure and temperature given.

$A_i=0.5m^2$ is the inlet area

$P_i=600Kpa$ pressure at the inlet area

$T_i=70C$ temperature at the inlet area

$A_o=1m^2$ is the outlet area

$P_o=100Kpa$ pressure at the outlet area

$T_o=C$ temperature at the outlet area

$\dot{m}=0.75\frac{kg}{s}$ represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

$\upsilon_i =0.04304\frac{kg}{m^3}$ and the entropy is $h_i=1.0645\frac{KJ}{KgK}=h_o$

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy $h_o=1.0645\frac{KJ}{KgK}$

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

$h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}$

$h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}$

Our interest value would be given using interpolation like this:

$\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}$

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that $\dot{m}=\frac{\dot{V}}{\upsilon}$ , but since $\dot{V}=Av$ we have this:

$\dot{m}=\frac{Av}{\upsilon}$

If we solve from the velocity v we have this:

$v=\frac{\upsilon \dot{m}}{A}$ (*)

And now we just need to replace the values into equation (*)

For the inlet case:

$v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}$

For the oulet case:

$v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}$

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3 years ago

A stoplight with weight 100 N is suspended at the midpoint of a cable strung between two posts 200 m apart. The attach points fo

Tasya [4]

There are 3 forces acting on the stoplight:

• its weight W, with magnitude W = 100 N, pointing directly downward

• two tension forces T₁ and T₂ with equal magnitude T₁ = T₂ = T = 1000 N, both making an angle of θ with the horizontal, but one points left and the other points right

The stoplight is in equilibrium, so by Newton's second law, the net vertical force acting on it is 0, such that

∑ F = T₁ sin(θ) + T₂ sin(180° - θ) - W = 0

We have sin(180° - θ) = sin(θ) for all θ, so the above reduces to

2T sin(θ) = W

2 (1000 N) sin(θ) = 100 N

sin(θ) = 0.05

θ ≈ 2.87°

If y is the vertical distance between the stoplight and the ground, then

tan(θ) = (15 m - y) / (100 m)

Solve for y :

tan(2.87°) = (15 m - y) / (100 m)

y = 15 m - (100 m) tan(2.87°)

y ≈ 9.99 m

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2 years ago

A rock climber stands on top of a 50 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s a

bagirrra123 [75]

<h2>Answer: b) What was the initial speed of the second stone?</h2>

Explanation:

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3 years ago