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3 years ago

Please show steps as to how to solve this problem Thank you!

Physics

1 answer:

bezimeni [28]3 years ago

4 0

Explanation:

Let x = distance of $F_1$ from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque $\tau_{net}$ about the fulcrum is zero:

$\tau_{net} = -F_1x + F_2d_2 = 0$

$-m_1gx + m_2gd_2 = 0$

$m_1x = m_2d_2$

Solving for <em>x</em>,

$x = \dfrac{m_2}{m_1}d_2$

$\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}$

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A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

The control group in an experimental study will __________.

Anettt [7]

Give you something to compare your results with. It's always nice to be able to see what changes have been made to the original, even if it's not technically the original (I know that was worded weird, I just don't know how else to explain it.) Hope this helped!

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3 years ago

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In order for a ball to move upward, the initial velocity of the ball must be greater than _____.

ch4aika [34]

Answer:

The answer is zero please Give me Brainly

Explanation:

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2 years ago

Work is done on a locked door that remains closed while you try to pull it open. True False.

dsp73

Answer:False

Explanation:

Work is being done on a body when it causes displacement of body on the application of force

$Work\ done=Force\times displacement$

When we pull the door by a force it causes zero displacements of the door. So we can say that work done on it is zero.

Thus the above-given statement is false

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3 years ago

A can of sardines is made to move along an x axis from x = 0.47 m to x = 1.20 m by a force with a magnitude given by F = exp(–8x

sattari [20]

If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .

I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect. On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on. The main question we're left with after all of this is: Why sardines ? ?

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