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almond37 [142]
2 years ago
9

A capacitor with initial charge q0 is discharged through a resistor. What multiple of the time constant τ gives the time the cap

acitor takes to lose
(a) the first 1/8-th of its charge
(b)7/8-th of its charge?
Physics
1 answer:
fiasKO [112]2 years ago
5 0

Answer: a) 0.13*τ ; b) 2.08*τ

Explanation: In order to explain the discharg of a capacitor through a resistor, we have to consider the following:

Q(t)=Qo* exp(-t/τ) for a lose of 1/8-th of its charge

in this case, Q(t)=7/8*Qo=7/8*exp(-t/τ)

ln(7/8)*τ=-t

then, t= -ln(7/8)*τ =0.13

For a lose of 7/8 th  of its charge , we have

Q(t)=1/7*Qo*exp(-t/τ)

t=-ln(1/8)*τ=2.08

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Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
2 years ago
The famous cliff divers of Acapulco leap from a perch 35 m above the ocean. How fast are they moving when they reach the surface
Rus_ich [418]

1) 26.2 m/s

The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:

E=K+U = \frac{1}{2}mv^2 + mgh

where

m is the mass of the diver

v is the speed

g = 9.8 m/s^2 is the acceleration due to gravity

h is the height above the water

When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:

E_i = mgh

where h=35 m is the height of the cliff.

When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:

E_f = \frac{1}{2}mv^2

since the total mechanical energy is conserved, we have

E_i = E_f\\mgh = \frac{1}{2}mv^2

And solving the equation for v, we find the speed when they reach the surface of the water:

v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(35 m)}=26.2 m/s

2) It is converted into thermal energy of the water

When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.

In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.

However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.

8 0
2 years ago
An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th
Aleksandr-060686 [28]
The sound is increased because sound waves are in fact mech. waves which means the  that they can't travel through empty space and thus need a medium to travel through
4 0
2 years ago
A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μs = 0.50 and μk = 0
frutty [35]

Answer:

x = 1.63 m

Explanation:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that most of the computers weight is applied on the belt instantaneously, we can apply the constant acceleration equation below

x = v^{2}/2a

where a = μk.g , therefore

x = v^{2}/2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m

8 0
2 years ago
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
2 years ago
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