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3 years ago

A particle moves along the x-axis so that its velocity at anytime is t greater than or equal to 0 is given by v(t)=1-sin(2pi t)

1 answer:

5 0

A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)

b) Any value of t can be plugged in as long as it is greater than or equal to 0.

c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π

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Answer with Explanation:

We are given that

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Direction: $\theta=tan^{-1}\frac{y}{x}$

By using the formula

Direction of D: $\theta=tan^{-1}(\frac{-2}{2})=tan^{-1}(-1)=tan^{-1}(-tan45^{\circ})=-45^{\circ}$

b.E=-A-B+C

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2 years ago

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Answer:

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1 year ago

a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a

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Answer:

$W = -2.76\times 10^{-9}~J$

Explanation:

The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.

The potential energy is given by the following formula:

$U = \frac{1}{2}CV^2$

where C can be calculated using the plate separation and area.

$C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon$

Therefore, the potential energy in the first case is

$U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon$

In the second case:

$C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon$

The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.

So, the difference in the potential energy is

$W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J$

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2 years ago

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Answer:

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Explanation:

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$a=\frac{v^2}{R}.$

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Therefore the acceleration is increased by a factor of 4.

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