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3 years ago

If we rub a rod and a fabric made of the same material, would the rod become charged? Please explain.

1 answer:

3 0

If you rub a glass rod with a cloth, the charge produced on the glass rod is called positive, whereas when a plastic rod is rubbed with a cloth, the charge produced on the plastic rod is negative.

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Please help... !!!!!!!

Scilla [17]

Answer:

1) C. The precise location of weather-related phenomena

2) C. It shows that the galaxies are moving farther away.

Explanation:

1) The global positioning system can be used in many ways, like in navigation, mapping and surveying the land. It is also used to observe the small changes taking place in the movement of objects on earth. They are usually used to study earthquakes.

2) When light from the galaxies is shifted towards the longer wavelengths, it is called red shift. When light shifts towards the shorter wavelengths, it is called blue shift. Big bang theory states how the primordial universe started expanding from a hot and dense singularity, in the process of evolution of the universe.

The universe is expanding always. Galaxies that are a part of the universe keep expanding and move farther away from each other. The galaxies that are very far off appear to move much faster than the nearby galaxies.

6 0

3 years ago

A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time

IrinaK [193]

Answer:

$8.0\mu C$

Explanation:

We are given that

$f=1.6 Hz$

$q=3.0\mu C=3.0\times 10^{-6} C$

$1\mu C=10^{-6} C$

Current,I= $75\mu A=75\times 10^{-6} A$

$1\mu A=10^{-6} A$

We have to find the maximum charge of the capacitor.

Charge on the capacitor, $q=q_0cos\omega t$

$\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s$

$3\times 10^{-6}=q_0cos3.2\pi t$ ....(1)

$I=\frac{dq}{dt}=-q_0\omega sin\omega t$

$75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t$ ....(2)

Equation (2) divided by equation (1)

$-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25$

$tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488$

$3.2\pi t=tan^{-1}(-2.488)=-1.188rad$

$q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C$

Hence, the maximum charge of the capacitor= $8.0\mu C$

4 0

4 years ago

A 13.3 kg box sliding across the ground

adelina 88 [10]

Answer:

0.25

Explanation:

According to newtons law of motion

\sum F_x = ma

F_f = ma

nR = ma

nmg = ma

ng = a

n = a/g

g is the acceleration due to gravity

Given

a = 2.42m/s²

g = 9.8m/s²

Substitute into the formula;

n = 2.42/9.8

n = 0.25

Hence the coefficient of kinetic friction is 0.25

5 0

3 years ago

A 23.0 kg child plays on a swing having support ropes that are 2.10 m long. A friend pulls her back until the ropes are 45.0 deg

Romashka [77]

Answer:

a. 139.748J

b.3.486m/s

c.zero

Explanation:

a. Given the mass of the child as 23.0kg, rope length is 2.1mand incline is 45°

Potential energy during release is calculated as: $PE=mgh$

#Find vertical difference of when the swing is at rest (2.1m) and when the child is pulled back.

Find the height when the child is pulled back:

$cos 45\textdegree=y/2.10\\\\y=1.48m$

#therefore,vertical difference is 2.1m-1.48m=0.62m

$\therefore PE=mgh\\\ \ =23.0kg\times 9.8m/s^2\times 0.62m\\\ \ =139.748J$

#Hence the potential energy during release is 139.748J

b. From a, above, we have PE=139.748J, M=23.0kg.

At the bottom, all the PE will be transferred into KE. Potential energy is calculated as:

$KE=0.5mv^2\\\\mv^2=2KE\\\\v=\sqrt{2KE/m}\\\\v=\sqrt{2\times 139.748J/23.0}\\\\v=3.486m/s$

#Hence the velocity at the bottom of the swing is 3.486m/s

c. Work is calculated as the product of force by distance.

From a, b above we have mass as 23.0kg .

-since the distance of the ropes remained constant the change in distance is zero:

$W=mgd\\=23.0\times 9.8m/s^2\times 0\\=0$

Therefore the work in the ropes is 0,zero.

8 0

3 years ago

A coil lies flat on a tabletop in a region where the magnetic field vector points straight up. The magnetic field vanishes sudde

Mademuasel [1]

Answer:

Option B - The induced current flows counter-clockwise.

Explanation:

Faraday's law of electromagnetic induction states that whenever a conductor is placed in a changing magnetic field, an electromotive force is induced and that if the conductor circuit is closed, a current is induced which is the induced current. The magnitude of the EMF induced in the coil is therefore proportional to the rate of change of magnetic flux throughout the coil.

Meanwhile, the direction of the induced current is given by Lenz's law which states that the direction of the induced current will oppose the change in electromagnetic force that produced that current.

Since the magnetic field points upwards, the induced current will move in a direction to the left which is counterclockwise

6 0

4 years ago