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3 years ago

In any problems involving circular motion, which way does the tangential speed vector point?

1 answer:

4 0

In what may be one of the most remarkable coincidences in
all of physical science, the tangential component of circular
motion points along the tangent to the circle at every point.

The object on a circular path is moving in that exact direction
at the instant when it is located at that point in the circle. The
centripetal force ... pointing toward the center of the circle ...
is the force that bends the path of the object away from a straight
line, toward the next point on the circle. If the centripetal force
were to suddenly disappear, the object would continue moving
from that point in a straight line, along the tangent and away from
the circle.

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The proper IUPAC naming convention for the compound symbol 'KI' is

Darya [45]

Proablyly a or c idk

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3 years ago

In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.

pickupchik [31]

In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because of friction.
AMA (Actual mechanical advantage) is found by dividing output force by effort force. The actual mechanical advantage will always be less than the ideal mechanical advantage. The ideal mechanical advantage assumes perfect efficiency which doesn't account for friction, while actual mechanical advantage does. Therefore; the IMA is always greater than the actual mechanical advantage because all machines must overcome friction.

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3 years ago

The electric force between two charged objects is influenced by

tresset_1 [31]

<span> attraction between the relative abundance of electrons in one object and protons in the other

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4 years ago

A closed circuit is formed by a solid conductor in contact with two parallel conducting rails that are closed on their left end,

castortr0y [4]

this is an equation that you need to solve for motional emf. motional emf=vBL, where v is velocity in meters/second, B is magnetic field in Teslas and L is length or distance the rails are apart from each other. when we plug everything into the formula given above, we get: motional emf=5m/s*0.80T*0.20m. solving all this we get 0.8 volts. pretty sure that since they are giving you the direction of the field, they want to know which way the current will flow . since the conductor is moving from left to right the area of the field is increasing which means magnetic flux is increasing as Ф(magnetic flux)=B(magnetic field)*A(area)*cosФ(little phi is the angle to the normal. in this case little fee is 0 degrees so the cosФ doesn't matter). so ↑Ф=B↑A. if magnetic flux is increasing, the induced magnetic field is in the opposite direction as the original magnetic field meaning the induced magnetic field will be out of the page. using the right hand rule which says that if the field is in to the page, the current should go clockwise and if the field is out of the page, the current is counterclockwise so that means that the current should be going counter clockwise since the induced field is going out of the screen. the top of the conducting wire will have its current go to the left and the bottom of the conducting wire will have the current go to the right.

8 0

3 years ago

An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr

notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

$Charge = Current \times Time$

As two different current is passing at two different times, the net charge will be the different in current. So,

$\text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t$

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

$V=\frac{k q}{R}$

Here $k = 9 * 10^9$ , q is the charge and R is the radius. As $q=2 \times 10^{-6} \times t$ and R =17 cm = 0.17 m, then the voltage will be

$V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

The time is required to find to reach the voltage of 1500 V, so

$1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}$

$\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}$

So, 14 ms is required to reach the potential of 1500 V.

3 0

3 years ago