Which two subatomic particles have approximately the same mass

If the current in a wire of a cd player is 12.0 ma, how long would it take for 6.50 c of charge to pass a point in this wire? an

Galina-37 [17]

The current intensity is defined as the amount of charge Q that passes through a certain point of an electrical wire in a time interval of $\Delta t$ :
$I= \frac{Q}{\Delta t}$

In our problem, the current intensity is
$I=12.0 mA=0.012 A$
while the amount of charge that passes a certain point of the wire is
$Q=6.50 C$

If we re-arrange the previous equation:
$\Delta t= \frac{Q}{I}$
we can find the time needed for this amount of charge to pass through a point of the wire:
$\Delta t= \frac{6.50 C}{0.012 A} =541.7 s$

6 0

3 years ago

A series RLC circuit with L = 12 mH, C = 3.5 mu or micro FF, and R = 3.3 ohm is driven by a generator with a maximum emf of 115

Elan Coil [88]

Explanation:

Given data

Inductance L=12*10^-³H

Capacitance C= 3.5*10^-6F

Resistance R= 3.3 Ohms

Voltage V=115v

Capacitive reactance Xc=?

inductive reactance Xl=?

Impedance Z=?

Phase angle =?

A. Resonance frequency

In RLC circuit resonance occurs when capacitive reactance equals inductive reactance

f=1/2pi √ LC

f=1/2*3.142 √ 12*10^-³*3.5*10^-6

f=1/6.284*0.0002

f=1/0.00125

f=800HZ

B. Find Irms at resonance.

Irms=R/V

Irms=3.3/115

Irms=0.028amp

Find the capacitive reactance XC in Ohms

Xc=1/2pi*f*C

Xc=1/2*3.142*800*3.5*10^-6

Xc=1/0.0176

Xc=56.8 ohms

To find the inductive reactance

Xl=2pifL

Xl=2*3.142*800*12*10^-3

Xl=60.3ohms

d) Find the impedance Z.

Z=√R²+(Xl-Xc)²

Z=√3.3²+(60.3-56.8)²

Z=√10.89+12.25

Z=√23.14

Z=4.8ohms

Phase angle =

Tan phi=Xc/R=56.8/3.3

Tan phi=17.2

Phi=tan-1 17.2

Phi= 1.51°

6 0

3 years ago

A ball is thrown up into the air with 100 j of kinetic energy, which is transformed to gravitational potential energy at the top

jenyasd209 [6]

The kinetic energy when it returns to its original height is 100 J

Solution:

The ball is thrown up with a Kinetic Energy K. E. = 0.5×m×v² = 100 J

Therefore the final height is given by

Where:

u = final velocity = 0

v = initial velocity

s = final height

Therefore v² = 2·g·s = 19.62·s

P.E = Potential Energy = m·g·s

Since v² = 2·g·s

Substituting the value of v² in the kinetic energy formula, we obtain

K. E. = 0.5×m×2·g·s = m·g·s = P.E. = 100 J

When the ball returns to the original height, we have

v² = u² + 2·g·s

Since u = 0 = initial velocity in this case we have

v² = 2·g·s and the Kinetic energy = 0.5·m·v²

Since m and s are the same then 0.5·m·v² = 100 J.

As the height of the ball increases the kinetic energy of the ball is converted into gravitational potential energy. This means that the kinetic energy of the bullet is reduced. When the ball reaches its maximum height, it momentarily comes to rest and the ball's kinetic energy is zero. When the ball hits the ground, its potential energy is converted to kinetic energy.

Learn more about Kinetic energy here:-brainly.com/question/25959744

#SPJ4

6 0

1 year ago

The atmosphere of earth does not escape into the space but remains attached to earth's surface why??

Alexeev081 [22]

Just like any other gas or mixture of gases, the gas molecules are
zipping around in all different directions and with a whole range of
different speeds.

Those that happen to be moving at a speed greater than the Earth's
"escape velocity", AND are pointed away from Earth, AND don't hit
any other molecules before they escape, are lost.

With the combination of Earth's escape velocity, and the temperatures,
thickness, and density of the atmosphere, that process happens slowly
enough to have maintained an atmosphere around this planet until now.

Personally, I hope it hangs around for a while longer. But with the constant
increase in temperature that's been going on, you never know . . .

7 0

3 years ago

Which of the following materials is an insulator against electric current

Leya [2.2K]

Aluminum foil (I think, it may not be right)

8 0

3 years ago

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