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Lera25 [3.4K]
2 years ago
13

Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50

0μf.
Physics
1 answer:
defon2 years ago
8 0

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

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A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu
Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0
2 years ago
7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr
wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho
Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)

80=5(50-x)

80=250-5x

5x=170

x=\frac{170}{5}

x=34 cm

4 0
2 years ago
A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t
Mashutka [201]
As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
f = (\frac{g + v_{r}}{g + v_{s}})f_{o} -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
v_{r} = Velocity of the receiver/observer relative to the medium = ?.
v_{s} = Velocity of the source with respect to medium = 0 m/s.
f_{o} =  Frequency emitted from source = 400 Hz.
f = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

408 = ( \frac{340 + v_{r}}{340 + 0})*400

\frac{408}{400} =  \frac{340 + v_{r}}{340}

Solving above would give you,
v_{r} = 6.8 m/s

The correct answer = 6.8m/s



7 0
2 years ago
A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f
snow_tiger [21]

Answer:

2.5 \times 10^{11} N-m upwards        

Explanation:

Torque is the vector cross product of the force and radial distance.

\tau = rF sin \theta

\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0
2 years ago
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