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2 years ago

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

1 answer:

7 0

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

t=0.476v
t=1.967v
V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

t = 1.0 seconds
5.0 seconds
20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

t=0.476v
t=1.967v
V2=4.323v

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trasher [3.6K]

Explanation:

It is given that,

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Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,

The current flowing through the inductor is given by :

$I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})$

I = 0.091 A

Energy stored in the inductor is, $U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2$

U = 0.000165 Joules

Hence, this is the required solution.

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3 years ago

Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o

spayn [35]

To solve letter a:

d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.

d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.

t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,

d = 115 * 0.71 = 81.65 km.

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2 years ago

All 2023 ariya ac synchronous drive motors produce ____% torque at 0 mph for impressive off-the-line acceleration and smooth cru

Savatey [412]

All 2023 Ariya ac synchronous drive motors produce 100% torque at 0 mph for impressive off-the-line acceleration and smooth cruising.

<h3>How to calculate the torque?</h3>

Mathematically, the torque of an automobile vehicle can be calculated by using this formula:

Torque = Fd

<u>Where:</u>

F is the force.

d is the distance.

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Read more on torque here: brainly.com/question/14839816

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2 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

$F_{racquet}=-2F_{player}$

Hence the magnitude of Force by racquet is twice the Force by player

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3 years ago

2NaOH+H2SO4 Na2SO4+H2O the equation is balanced/unbalanced because the number of hydrogen atoms and oxygen/sodium is equal/not e

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Your equation is:
$2NaOH + H_{2}SO_{4} ==\ \textgreater \ Na_{2}SO_{4} + H_{2}O$

An equation is balanced only if there are the same number of atoms of each element on both sides of the arrow - aka same number of atoms of each element in both reactants (left of the arrow) and products (right of the arrow).

It'll be easiest to tackle this by counting up the number of atoms of each element on the left and on the right and comparing those numbers. If there is a number in front of the entire compound, that means that number applies to all elements in the compound. If the number is a subscript (little number to the right of the element), that means that number only applies to the element that the subscript is attached to:
1) On the left, you have:
$2NaOH + H_{2}SO_{4}\\
\\
Na: \: 2 \: atoms\\
O: \: 2 + 4 = 6 \: atoms\\
H: \: 2 + 2 = 4 \: atoms\\
S: \: 1 \: atom$

2) On the right, you have:
$Na_{2}SO_{4} + H_{2}O\\
\\ 
Na: \: 2 \: atoms\\ 
O: \: 4 + 1 = 5 \: atoms\\ 
H: \: 2 \: atoms\\ 
S: \: 1 \: atom$

You can see that the number of oxygen and hydrogen atoms aren't equal on both the left (reactants) and the right (products), so the equation is unbalanced.

Your final answer is "T<span>he equation is unbalanced because the number of hydrogen atoms and oxygen is not equal in the reactants and in the products."</span>

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3 years ago

Read 2 more answers