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3 years ago

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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:

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
$L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}$

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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:

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

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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
$W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J$

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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.

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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 $F=kA=290400\times 0.1=29040N$

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3 years ago