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il63 [147K]
3 years ago
11

In an RC series circuit, ε = 12.0 V, R = 1.25 MΩ, and C = 1.42 µF. (a) Calculate the time constant. (b) Find the maximum charge

that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 8.78 µC?
Physics
1 answer:
faust18 [17]3 years ago
8 0

Answer:

a, 1.775s

b, 17.04μC

c, 1.28s

Explanation:

Given

R = 1.25MΩ

C = 1.42µF

ε = 12.0 V

q = 8.78 µC

Time constant, τ = RC

τ = (1.25*10^6) * ( 1.42*10^-6)

τ = 1.775s

q• = εC

q• = 12 * 1.42*10^-6

q• = 17.04*10^-6C

q• = 17.04μC

Time t =

q = q• [1 - e^(t/τ)]

t = τIn[q•/(q•-q)]

t = 1.775In[17.04μC/(17.04μC-8.78μC)]

t = 1.775In(2.06)

t = 1.775*0.723

t = 1.28s

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A 1000kg car is rolling slowly across a level surface at 1 m/s heading twoards a group o fsmall innocent children. The doors are
Degger [83]

Answer:

The force required to push to stop the car is 288.67 N

Explanation:

Given that

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, \theta=30^{\circ}

Distance, d = 2 m

Let F is the force must you push to stop the car.

According work energy theorem theorem, the work done is equal to the change in kinetic energy as :

W=\dfrac{1}{2}m(v^2-u^2)F\times d=\dfrac{1}{2}m(v^2-u^2)

v = 0

Fd\ cos\theta=\dfrac{1}{2}m(u^2)      F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}F = -288.67 N

The force required to push to stop the car is 288.67 N

3 0
3 years ago
Before being engulfed, matter that is pulled into a black hole should become very hot and emit _____.
miss Akunina [59]
A black hole is a cosmological object that is created when a massive star comes to the end of its life and collapses under its own gravity. Black holes have massive gravitational fields  that even light cannot escape beyond a certain distance. Before being engulfed, matter that is pulled into a black hole should become very hot and emit electromagnetic radiation. 
4 0
4 years ago
Read 2 more answers
When a piece of iron or mass 78 gm is put in a graduated cylinder containing 100 cm cube of water the reading of the cylinder be
Wittaler [7]

Answer:

Ro = 7.8 [g/cm³]

Explanation:

According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.

V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]

We now know that density is defined as the relationship between mass and volume.

Ro = m/V_{d}

where:

Ro = density [g/cm³]

m = mass = 78 [g]

Vd = displacement volume [cm³]

Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]

7 0
3 years ago
Light travels through a liquid at 2.25e8 m/s. What is the index of refraction of the liquid?
Pavlova-9 [17]

Answer:

The index of refraction of the liquid is n = 1.33 equivalent to that of water

Explanation:

Solution:-

- The index of refraction of light in a medium ( n ) determines the degree of "bending" of light in that medium.

- The index of refraction is material property and proportional to density of the material.

- The denser the material the slower the light will move through associated with considerable diffraction angles.

- The lighter the material the faster the light pass through the material without being diffracted as much.

- So, in the other words index of refraction can be expressed as how fast or slow light passes through a medium.

- The reference of comparison of how fast or slow the light is the value of c = 3.0*10^8 m/s i.e speed of light in vacuum or also assumed to be the case for air.

- so we can mathematically express the index of refraction as a ratio of light speed in the material specified and speed of light.

- The light passes through a liquid with speed v = 2.25*10^8 m/s :

                         n = c / v\\\\n = \frac{ 3*10^8 }{2.25*10^8} \\\\n = 1.33

- The index of refraction of the liquid is n = 1.33 equivalent to that of water.    

         

8 0
3 years ago
Read 2 more answers
Hey!
pav-90 [236]

Answer:

Hydraulic pressure exerted on glass slab, ρ=10 atm

Bulk modulus of glass, B=37×10^9 Nm^−2

Bulk modulus, B=P/(ΔV/V)

where,

ΔV/V= Fractional change in volume

ΔV/V=P/B

=10×1.013×10^5 /(37×10 ^9)

=2.73×10^-5

Therefore, the fractional change in the volume of the glass slab is 2.73×10^-5

Hope it helps

3 0
2 years ago
Read 2 more answers
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