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2 years ago

Explain the difference between mass and weight and how they are measured.

Physics

1 answer:

tamaranim1 [39]2 years ago

8 0

Answer:

Mass is a measure of how much force it will take to change that path. Mass depends on how much matter – atoms and so on – there is in an object; more mass means more inertia, as there is more to get moving. ... Weight, on the other hand, is a measure of the amount of downwards force that gravity exerts on an object

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What is the amplitude of an AC voltage waveform, in units of Volts, if the RMS value is 369 V?

Blizzard [7]

Answer:

521.8 V

Explanation:

The RMS value of the voltage of an AC signal is given by

$V_{rms}=\frac{V_0}{\sqrt{2}}$

where

$V_0$ is the peak voltage, which corresponds to the amplitude of the AC waveform

In this problem, we know the RMS voltage

$V_{rms}=369 V$

Therefore, we can re-arrange the previous equation to find the peak voltage (the amplitude of the waveform):

$V_0 = \sqrt{2}V_{rms}=\sqrt{2}(369 V)=521.8 V$

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2 years ago

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2 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.

3 0

2 years ago

If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini

kondaur [170]

<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>

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2 years ago

One hypothesis states that plate movement results from convection currents in the

brilliants [131]

I believe is A) Inner core

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3 years ago