1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
alex41 [277]
3 years ago
10

A capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280-mH induc

tor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time (the moment of connection with the inductor);
Physics
1 answer:
ololo11 [35]3 years ago
4 0

Frequency = 1 / (2π√LC)

Frequency = 1 / (2π · 2.8 x 10⁻⁴ · C)

Frequency = 1 / (1.759 x 10⁻³ · C)

<em>Frequency = (568.4 / C)  Hz.</em>

<em></em>

Energy stored in a capacitor = 1/2 C V²

Energy = 1/2 C · 2.25 x 10⁴

<em>Energy = (11,250 · C)  Joules</em>

<em></em>

Neither of the answers can be completely specified without knowing the value of the capacitor.

You might be interested in
A car travelling at 32 m/s decelerates at a rate of 6 m/s ^ 2 over 4 secondsWhat is the final velocity the car
Pachacha [2.7K]

Answer:

v=8m/s

Explanation

The initial velocity, u, of the car=32m/s

Since the car has decelerated, we negate the acceleration

This implies that,

a=-6m{s}^{ -2}

Time,t=4s

Let v be the final velocity,

using the relation,

a =  \frac{v - u}{t}

We obtain the below by substitution

- 6= \frac{v - 32}{4}

By cross multiplying we obtain,

- 6 \times 4 = v - 32

- 24=v-32

- 24+32=v

v=8m/s

5 0
3 years ago
For an unknown sample of the experiment, students measure 1679 counts when they first receive their sample and 1336 counts four
Tpy6a [65]

Answer:

Half life of the sample, t_{\frac{1}{2}} = 12.15 min

Given:

Initial amount, N = 1679

Final count of amount, N' = 1336

Time elapsed, t = 4 min = 240 s

Solution:

Now, To calculate the half life, using the relation:

N' = N(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}

Now, substituting the given values in the above mentioned formula:

\frac{1336}{1679} = (\frac{1}{2})^{\frac{4}{t_{\frac{1}{2}}}}

0.796 = 0.5^{\frac{4}{t_{\frac{1}{2}}}}

Taking log on both the sides:

ln(0.796) = \frac{4}{t_{\frac{1}{2}}}ln(0.5)

0.329 = 4t_{\frac{1}{2}}

t_{\frac{1}{2}} = 12.15 min

6 0
4 years ago
A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed
AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

\omega=\sqrt{\frac{k}{m}}

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s

And the frequency of the motion instead is given by:

f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

F=kA

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N

k = 160 N/m is the spring constant

Solving for A, we find

A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m

3 0
3 years ago
Fill in the blanks:
Dafna11 [192]

Answer:

st8tif7u

Explanation:

t996w4sia74a85e8srz

7 0
3 years ago
When a hot object is placed in a water bath whose temperature is 25◦C, it cools from 100 to 50◦C in 150 s. In another bath, the
mixas84 [53]
Using Newton's Law of Cooling, the given are:
Temperature of the first bath (Ta1): 25C
a(final temperature)= 50C, b (initial temperature)=100C
x-y is the difference in time. In the first bath, 150s

dT/dt=-k(T-Ta)
\int\limits^a_ b{(dT/(T-Ta1))} \, dt = \int\limits^x_y{-k} \, dt
ln (T-25) [from a=100 to b=50]= -k*150
ln ((50-25)/(100-25))= -k*150

Solving for k=7.324081X10^-3

For the second bath: find Ta2 with dt=120s

dT/dt=-(7.324081x10^-3)*(T-Ta2)
\int\limits^a_ b{(dT/(T-Ta2))} \, dt = \int\limits^x_y{-7.324081x10^-3} \, dt
ln (T-Ta2) [from a=100 to b=50]= -(7.324081x10^-3)*120
ln ((50-Ta2)/(100-Ta2))= -(<span>7.324081x10^-3)</span>*120

Solving for Ta (second bath temperature)= 14.49 degrees Celsius

6 0
3 years ago
Read 2 more answers
Other questions:
  • 1. Use the diagram to anwser question 1.
    6·2 answers
  • What are the three points of the fire triangle
    6·1 answer
  • Finding Impulse when u hav force and time
    12·1 answer
  • The acceleration of an object depends upon
    11·2 answers
  • Referring to the diagram above, predict what will happen when the switch is closed. Explain your answer.
    5·1 answer
  • From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ
    15·1 answer
  • Why do astronauts' weights differ in space?
    13·1 answer
  • Suppose you have two metal spheres that are exactly the same size, separated by a very large distance. Sphere A carries an exces
    8·1 answer
  • A lamp draws a current of 20 A when it is connected to a 100 V source? What is the resistance of the lamp?
    11·1 answer
  • How can I convert this?<br>Please answer with solution. Thank you.​
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!