Which of the following describes a referee's job?

A simple pendulum has a bob of mass M. The bob is on a light string of length . The string is fixed at C. At position A, the str

Vlad [161]

Answer:

$v=\sqrt{2gL}$

Explanation:

mass of bob = M

string is fixed at C, at position A the string is horizontal and at position B teh string is vertical.

Let the length of the string is L.

At the point C, it has maximum potential energy which is equal to

U = M x g x L ..... (1)

At the position B, it has maximum kinetic energy and the velocity is v.

K = 1/2 Mv² ...... (2)

According to the conservation of energy

The potential energy at the position A is equal to the kinetic energy at position B.

M x g x L = 1/2 M x v²

v² = 2 x g x L

$v=\sqrt{2gL}$

6 0

3 years ago

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The low-frequency speaker of a stereo set has a surface area of and produces 1W of acoustical power. What is the intensity at th

AlekseyPX

Answer:

I = $\frac{1}{4\pi \ r^2}$

we see the intensity decreases with the inverse of the distance squared

Explanation:

Intensity is defined as power per unit area,

I = P / A

in this case we have that the sound is emitted in a spherical form therefore the area is

A = 4 pi r2

therefore the intensity is

I = $\frac{1}{4\pi \ r^2}$

as we see the intensity decreases with the inverse of the distance squared

5 0

3 years ago

Find p1, the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.)

Taya2010 [7]

Answer:

(a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Explanation:

Given that,

Gauge pressure at bottom = p₁

Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.

Find the speed of the fluid in the left end of the main pipe.

(a). We need to calculate the gauge pressure at the bottom of tube 1

Using bernoulli equation

$P_{1}=\rho g h_{1}$

(b). We need to calculate the speed of the fluid in the left end of the main pipe

Using bernoulli equation

Pressure for first pipe,

$P_{1}=\rho gh_{1}$ .....(I)

Pressure for second pipe,

$P_{2}=\rho gh_{2}$ .....(II)

From equation (I) and (II)

$P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

Put the value of P₁ and P₂

$\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)$

$2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2$ ....(III)

We know that,

The continuity equation

$v_{1}A_{1}=v_{2}A_{2}$

$v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})$

Put the value of v₂ in equation (III)

$2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2$

$2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2$

Here, $\dfrac{A_{1}}{A_{2}}=\gamma$

So, $2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)$

$v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

Hence, (a). The gauge pressure at the bottom of tube 1 is $P_{1}=\rho g h_{1}$

(b). The speed of the fluid in the left end of the main pipe $\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}$

8 0

3 years ago

The chart below lists six stars. Research the luminosity, distance from Earth, and surface temperature of each star. Make sure t

goldenfox [79]

I think is true but not sure

6 0

3 years ago

Which statement best describes the density of the outer planets?

gavmur [86]

Answer:

Saturn is mainly composed of the lightest two gases known, hydrogen and helium. It is the only planet in our solar system whose density is less than water.

Explanation:

5 0

4 years ago

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