PHYSICS COLLEGE

Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

Answers

Answer 1

Answer:

Answer:

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

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A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

Answers

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

h = 16.9 m

Which of these 23rd chromosomecombinations is likeliest to result in a
person with male and female traits?
ΧΟ
XXX
XXY
XY

Answers

Sorry if I’m wrong but I think it’s XO since o is not a sex chromosome

Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025 kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)? h2 / h1 =

Answers

Answer:

Explanation:

Given

density of cylinder is $\rho _c=713 kg/m^3$

Length of first cylinder is $L_1=20 cm$

radius $r_1=5 cm$

For cylinder 2 $L_2=10 cm$

$r_2=10 cm$

$h_1$ and $h_2$ are the height above water

as object is floating so its weight must be balanced with buoyant force

$\rho _c(\pi )/(4)d_1^2L_1g=\rho _w(\pi )/(4)d_1^2(L_1-h_1)g----1$

For 2nd cylinder

$\rho _c(\pi )/(4)d_2^2L_2g=\rho _w(\pi )/(4)d_2^2(L_2-h_2)g----2$

Dividing 1 and 2 we get

$(L_1)/(L_2)=(L_1-h_1)/(L_2-h_2)$

$(20)/(10)=(20-h_1)/(10-h_2)$

$2h_2=h_1$

$\n\Rightarrow(h_2)/(h_1)=(1)/(2)$

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/s

E. None of the option is correct

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

What are supersonic speeds

Answers

I think your answer is speed faster than the speed of sound

Answer:

speeds above 343 m/s

Explanation:

I have taken the test got 100%

Which type of diffraction occurs when the point source and the screen are at finite distances from the obstacle forming the diffraction pattern?A. fraunhofer
B. fresnel
C. far-field
D. single slit

Answers

Fresnel diffraction

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