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2 years ago

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Give one example where the displacement is zero but the distance traveled is not zero.

Physics

1 answer:

andriy [413]2 years ago

5 0

Let us assume a man travelled from a point A to a point B over a distance 'd'. After a while he travels the same distance back to point A.

Therefore, since his initial and final positions are the same, displacement is equal to zero, and the distance travelled is (d + d) = 2d, which is not zero.

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A small object is attached to the end of a relaxed, horizontal spring whose opposite end is fixed. The spring rests on a frictio

eduard

Answer:4A

Explanation:

Given

Mass is displace x= A units from its mean position x=0'

When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A

One cycle is completed when block returns to its original position

so first block will go equilibrium position x=0 and then to x=-A

from x=-A it again moves back to x=0 and finally back to its starting position x=A

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3 years ago

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.

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WILL GIVE BRAINLY!! PLEASE HELPP

LekaFEV [45]

The purpose of the machine is to leverage its mechanical advantage such that the force it outputs to move the heavy object is greater than the force required for you to input.

But there's no such thing as a free lunch! When you apply the conservation of energy, the work the machine does on the object will always be equal to (in an ideal machine) or less than the work you input to the machine.

This means that you will apply a lesser force for a longer distance so that the machine can supply a greater force on the object to push it a smaller distance. That is the trade-off of using the machine: it enables you to use a smaller force but at the cost of having to apply that smaller force for a greater distance.

The answer is: The work input required will equal the work output.

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The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would pro

Gnesinka [82]

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of <u>Eddington luminosity</u>

<h3>What are stars?</h3>

Stars are a fixed luminous point in the sky which is a large and remote incandescent body

So therefore, the hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity

Learn more about stars:

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<span>more lines = a lot of electrons returning back to ground state from same level</span>

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