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s344n2d4d5 [400]

2 years ago

15

What's the answer for question 2?

1 answer:

Luden [163]2 years ago

8 0

Sway bar ??? Idk if that's right

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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant

atroni [7]

Answer:

The time constant is $\tau = 1.265 s$

Explanation:

From the question we are told that

the time take to charge is $t = 2.4 \ s$

The mathematically representation for voltage potential of a capacitor at different time is

$V = V_o - e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant

$V_o$ is the potential of the capacitor when it is full

So the capacitor potential will be 100% when it is full thus $V_o =$ 100% = 1

and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

V = 0.85

Hence

$0.85 = 1 - e^{-\frac{2.4}{\tau } }$

$- {\frac{2.4}{\tau } } = ln0.15$

$\tau = 1.265 s$

7 0

2 years ago

You have What needs to be done to this circuit so that the light bulb lights up?

Fynjy0 [20]

Close the switch would be the correct answer

6 0

2 years ago

Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =

Airida [17]

Answer:

If the final question is; at what velocity will the first block start to move outward in m/s?

$v = 3.5596 \frac{m}{s}$

Explanation:

The motion have the velocity that will make the block move using:

$F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}$

$m_{1} = 1.2 Kg$

$m_{2} = 2.8 Kg$

$r = 0,45 m$

μ $s= 0,54$

Resolving:

$\frac{m_{1}*v^{2} }{r} = us*m_{1}* g + m_{2} * g$

$v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }$

$v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}$

$v^{2} = 12.6714 \frac{m^{2} }{s^{2} }$

$v = 3.559 \frac{m}{s}$

4 0

2 years ago

A proton traveling with speed 2 × 105 m/s in the -y direction passes through a region in which there is a uniform magnetic field

stepladder [879]

Answer:

$\vec{E} = 1.2\times 10^5(-i)$

Explanation:

Given that

Speed ,v= 2 x 10⁵ m/s ( - y direction)

B= 0.6 T (- z direction)

The resultant force on the proton given as

$\vec{F}=q.\vec{E}+ q.(\vec{v}\times \vec{B})$

F= m a

For uniform motion acceleration should be zero.

F = 0

$0=q.\vec{E}+ q.(\vec{v}\times \vec{B})$

$0=\vec{E}+ (\vec{v}\times \vec{B})$

$0=\vec{E}+2\times 10^5(-j) \times 0.6(-k)$

$\vec{E} =- 2\times 10^5(-j) \times 0.6(-k)$

$\vec{E} =-1.2\times 10^5(i)$

$\vec{E} = 1.2\times 10^5(-i)$

Electric filed should be apply in the negative x direction.

5 0

2 years ago

CAN SOMEONE PLZ TELL ME HOW TO WRITE THIS DATA IN THE THE TABLE INTO A GRAPH?

vitfil [10]

Here is a helpful video https://www.khanacademy.org/_render

6 0

2 years ago