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Rom4ik [11]
3 years ago
9

What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Physics
1 answer:
12345 [234]3 years ago
5 0

Incomplete question.The complete question is

A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder? Suppose that the density of oil is 900 kg/m3.

Answer:

Total gauge pressure= 3724Pa or 3.724kPa

Explanation:

Pressure (water alone) = (1,000 x 9.8 x 0.2m) = 1,960Pa.

Pressure (oil on top) = (900 x9.8 x 0.2m) = 1,764Pa.

Total gauge pressure at bottom = (1,960 + 1,764) = 3724Pa or 3.724kPa

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A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov
Brums [2.3K]

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J

6 0
3 years ago
Which of the following options is correct and why?
Dimas [21]

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

6 0
3 years ago
Match the indices of refraction with the corresponding effects on the waves.
Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the  speed of light in a vacuum and the speed of light in that medium:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

v=\frac{c}{n}

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1 ,n_2 are the index of refraction of the 1st and 2nd medium

\theta_1,\theta_2are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

So we see that if the index of refraction of the second medium is lower (n_2), then the ratio \frac{n_1}{n_2} is larger than 1, so the angle of refraction is larger than the angle of incidence:

\theta_2>\theta_1

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

3 0
3 years ago
Suppose we have two planets with the same mass, but the radius of the second one is twice the size of the first one. How does th
bogdanovich [222]

The free-fall acceleration on the second planet is one-fourth the value of the first planet.

Calculation:

Consider the mass of planet A to be, M

               the mass of planet B to be, Mₓ = M

               the radius of planet A to be, R₁

               the radius of planet B to be, R₂

The acceleration due to gravity on planet A's surface is given as:

g = GM/R₁²      - (1)

Similarly, the acceleration due to gravity on planet B's surface is given as:

g' = GM/R₂²                           [where, R₂ = 2R₁]

   = GM/4R₁²    -(2)

From equation 1 & 2, we get:

g/g' = GM/R₁² ÷ GM/4R₁²

g/g' = 4/1

Thus we get,

g' = 1/4 g

Therefore, the free-fall acceleration on the second planet is one-fourth the value of the first planet.

Learn more about free-fall here:

<u>brainly.com/question/13299152</u>

#SPJ4

6 0
2 years ago
Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and plac
mylen [45]

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

W = U_f - U_i

Where,

W = Work

U_f = Final gravitational Potential energy

U_i = Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

U_i = \frac{-GMm}{R}

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is R_{orbit} \approx R then the energy possessed by the spacecraft is

U_f = \frac{-GMm}{2R}

Work needed to put it in orbit is the difference between the above two

W = U_f - U_i

W = -GMm (\frac{1}{2R}-\frac{1}{R})

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

W = \frac{GMm}{2R}

3 0
3 years ago
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