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Ulleksa [173]
2 years ago
14

Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener

gy kb of particle
b. remember that both particles rotate about the y axis.
Physics
1 answer:
Flauer [41]2 years ago
5 0
<span>Answer: KE = (11/2)mω²r², particle B must have mass of 2m, while A has mass m. Then the moment of inertia of the system is I = Σ md² = m*(3r)² + 2m*r² = 11mr² and then KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2 So I'll proceed under that assumption. For particle A, translational KEa = ½mv² but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r² For particld B, translational KEb = ½(2m)v² but v = ω*r, so KEb = ½(2m)ω²r² so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2 which is equal to our rotational KE.</span>
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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
Drag the tiles to the correct boxes to complete the pairs. Match the sentences with the steps of the scientific method
Assoli18 [71]

Solution:

Make an Observation - An indoor plant in a dark room withers faster than the same plant in a room with ample sunlight.

Ask a question- Why do certain indoor plants die faster based on where they are placed in the house?

State a hypothesis- Sunlight is probably essential for plants to grow and live.

Run an experiment- Get two potted plants. Cover one with black paper. Place both plants outside in sunlight. See what happens to each plant after one week.

Analyze the results-The plant in the pot with black paper withered. The other plant was healthy.

Communicate the results to others - Plants need sunlight to make food so they can live.

4 0
2 years ago
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
Felectric = q*E 
<span> Ftranslational = m*a 
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
</span><span> Solve for a 
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
</span><span> m = 1.67*10^-27 kg 
</span><span> E = -150 N/C (- because it is down and up is positive) 
</span> a =<span> -6,4*10^5</span><span> m/s^2 (downward) 
</span> answer
 a = -6,4*10^5 m/s^2 (downward) 
3 0
2 years ago
A solid cylinder of mass 12.0 kg and radius 0.250 m is free to rotate without friction around its central axis. If you do 75.0 J
faltersainse [42]

Answer:

20 rad/s

Explanation:

mass, m = 12 kg

radius, r = 0.250 m

Moment of inertia of cylinder, I = 1/2 mr²

I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2

Work done = Change in kinetic energy

Initial K = 0

Final K = 1/2 Iω²

W = 1/2 Iω²

ω² = 2W/ I = 2 x 75 / (0.375)

ω = 20 rad/s

Thus, the final angular velocity is 20 rad/s .

8 0
2 years ago
Astronomers initially had difficulty identifying the emission lines in quasar spectra at optical wavelengths because
Rus_ich [418]
No one expected violet & ultraviolet spectral lines to be shifted towards the red.
6 0
2 years ago
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