1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
sdas [7]
3 years ago
15

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

m of the stone about the center of the circle is:
Physics
1 answer:
Illusion [34]3 years ago
3 0
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
You might be interested in
A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at
Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

        m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)

  • Rearranging terms, we have:

        m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)

  • We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

        \Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1}  *v_{1f} ^{2}  + \frac{1}{2}* m_{2}  *v_{2f} ^{2} (3)

  • Rearranging , and simplifying common terms, we have:

        m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2}  *v_{2f} ^{2} (4)

  • Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

        v_{10} + v_{1f} = v_{2f}

  • Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

       m_{1} * v_{10}  = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3}  = \frac{-17.1cm/s}{3} = -5.7 cm/s

7 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
Suppose that a car skids 15 meters if it is moving at 50 kilometers per hour Acceleration-Velocity-Position when the brakes are
julia-pushkina [17]

Explanation:

Below is an attachment containing the solution.

4 0
2 years ago
Simone created a chart to summarize the energy transformations that take place when energy from the wind is used to generate ele
melomori [17]

Answer:

d) kinetic energy transformed to mechanical energy

Explanation:

Wind energy comes from its movement, so kinetic energy

         Em = K = ½ m v²2

This energy spins the mill aspadle that this movement of the rotor within a magnetic field creates electricity in accordance with Faraday's law.

Consequently, from the above we should make a graph of the wind speed (kinetic energy) according to the electricity produced

The correct answer is d

4 0
3 years ago
Read 2 more answers
The motion of a piston in an auto engine is simple harmonic. If the piston travels back and forth over a distance of 10 cm and t
chubhunter [2.5K]

Answer:

Maximum force will be 29040 N

Explanation:

We have given mass of the piston m = 1.5 kg

Amplitude A = 10 cm = 0.1 m

Angular speed \omega =4200rpm=\frac{2\times 3.14\times 4200}{60}=440rad/sec

We know that angular speed is given by \omega =\sqrt{\frac{k}{m}}

440=\sqrt{\frac{k}{1.5}}

Squaring both side

193600=\frac{k}{1.5}

k = 290400 N/m

Now force is given by F=kA=290400\times 0.1=29040N

3 0
3 years ago
Other questions:
  • A woman flies 300 kilometers north, 200 kilometers west, and 500 kilometers south. How far is she from her starting point?
    7·1 answer
  • After a projectile is fired into the air, what is the acceleration in the x-direction? (Assume no air resistance.)
    7·2 answers
  • 1. Johnny wants to know where the water line is in a dark well. He drops a penny into the well and counts until he hears the pen
    11·1 answer
  • The Gulf Stream is an ocean current that runs from the southern tip of Florida up the eastern coast of the U.S. And ends in the
    7·1 answer
  • What is the difference between Helium3, and Helium?
    14·1 answer
  • A train accelerates from rest and covers a distance of 600m in 20s. What is the train's speed at the end of 20s? What is the fin
    5·1 answer
  • Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!
    12·1 answer
  • The positron created in pair production travels a certain distance and loses all of its kinetic energy. It finally annihilates w
    9·1 answer
  • Two children are riding on a merry-go-round. Child A is at a greater distance from the axis of rotation than child B. Which chil
    13·1 answer
  • Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!