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2 years ago

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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.

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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Complete Question:

Football player A has a mass of 110 kg, and he is running down the field with a velocity of 2 m/s. Football player B has a mass of 120 kg and is stationary. What is the total momentum after the collision?

Answer:

Total momentum = 220 Kgm/s.

Explanation:

<u>Given the following data;</u>

For footballer A

Mass, M1 = 110kg

Velocity, V1 = 2m/s

For footballer B

Mass, M1 = 120kg

Velocity, V1 = 0m/s since he's stationary.

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

a. To find the momentum of A;

$Momentum \; A = 110 * 2$

Momentum A = 220 Kgm/s.

b. To find the momentum of B;

$Momentum \; B = 120 * 0$

Momentum B = 0 Kgm/s.

c. To find the total momentum of the two persons;

$Total \; momentum = Momentum \; A + Momentum \; B$

Substituting into the equation, we have;

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Work may be measured using units of <br> A.watts<br> B.newton<br> C.joules<br> D.newtons per second

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3 years ago

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 18.0 m/s. If the soccer ball has a mas

Radda [10]

Answer:

change in momentum, $\Delta p=7.65 \,kg.m.s^{-1}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

Average Force, $F=144.3396\,N$

$F_x=125.0018\,N$

$F_y=72.1698\,N$

Explanation:

Given:

angle of kicking from the horizon, $\theta= 30^{\circ}$

velocity of the ball after being kicked, $v=18 m.s^{-1}$

mass of the ball, $m=0.425\, kg$

time of application of force, $t=5.3\times 10^{-2}\,s$

We know, since body is starting from the rest

$\Delta p=m.v$ .....................(1)

$\Delta p=0.425\times 18$

$\Delta p=7.65 \,kg.m.s^{-1}$

Now the components:

$\Delta p_x= 7.65\times cos 30^{\circ}$

$\Delta p_x= 6.6251 \,kg.m.s^{-1}$

similarly

$\Delta p_y= 7.65\times sin 30^{\circ}$

$\Delta p_y= 3.825 \,kg.m.s^{-1}$

also, impulse

$I=F\times t$ .........................(2)

where F is the force applied for t time.

Then from eq. (1) & (2)

$F\times t=m.v$

$F\times 5.3\times 10^{-2}= 7.65$

$F=144.3396\,N$

Now, the components

$F_x=144.3396\times cos 30^{\circ}$

$F_x=125.0018\,N$

&

$F_y=144.3396\times sin 30^{\circ}$

$F_y=72.1698\,N$

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2 years ago

A flywheel of J = 50 kg-m2 initially standing still is subjected to a constant torque. If the angular velocity reaches 20 Hz in

Karolina [17]

Answer:

$\tau = 1256.5\ N.m$

Explanation:

given,

J = 50 kg-m²

frequency, f = 20 Hz

time ,t = 5 s

we know,

angular velocity = 2 π f

ω = 2 π x 20

ω = 125.66 rad/s

now, angular acceleration calculation

$\alpha = \dfrac{\omega_f-\omega_i}{t}$

$\alpha = \dfrac{125.66-0}{5}$

α = 25.13 rad/s²

Torque given to the flywheel.

$\tau = I \alpha$

$\tau = 50\times 25.13$

$\tau = 1256.5\ N.m$

Torque of the given flywheel is equal to $\tau = 1256.5\ N.m$

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