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3 years ago

6

I. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maxi

mum compression. Indicate whether the work done by the spring is positive, negative, or zero.

1 answer:

nalin [4]3 years ago

6 0

Answer:

Negative

Explanation:

If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.

Hope this helped. :)

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A series circuit has a capacitor of 10−5 F, a resistor of 3 × 102 Ω, and an inductor of 0.2 H. The initial charge on the capacit

Alexus [3.1K]

Given the values to proceed to solve the exercise, we resort to the solution of the exercise through differential equations.

The problem can be modeled through a linear equation, in the form:

$10^5 Q +300Q'+0.2Q''=0$

With the initial conditions as,

$Q(0) = 10^{-6}$

$Q'(0)= 0$

Where Q(t) is the charge.

<em>The general solution of a linear equation is given as:</em>

<em> $y(x) = c_1e^{-ax}+c_2e^{-bx}$ </em>

Applying this definiton in our differential equation we have that

$Q(t) = C_1e^{at}+C_2e^{bt}$

To find b and a we use the first equation and find the roots:

$r_{a,b} = \frac{-300 \pm \sqrt{(300)^3-4(0.2)*10^5}}{0.4}$

$r_{a,b} = {-1000,-500}$

Then we have

$Q(t) = C_1e^{-1000t}+C_2e^{-500t}$

To find the values of the Constant we apply the initial conditions, then

$Q(0)= 10^{-6} = C_1+C_2$

And for the derivate:

$Q'(t) = -1000C_1e^{-1000t}-500C_2e^{-500t}$

$0 = -1000C_1e^{-1000(0)}-500C_2e^{-500(0)}$

$0 = -1000C_1-500C_2$

We have a system of 2x2:

$(1) 10^{-6} = C_1+C_2$

$(2) 0 = -1000C_1-500C_2$

Solving we have:

$C_1 = -10^{-6}$

$C_2 = 2*10^{-6}$

The we can replace at the equation and we have that the Charge at any moment is given by,

$Q(t) = (-10^{-6})e^{-1000t}+( 2*10^{-6})e^{-500t}$

If we obtain the derivate we find also the Current, then

$I(t)= 10^{-3}e^{-1000t}-10^{-3}e^{-500t}$

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3 years ago

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Naddik [55]

Answer:

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Explanation:

Data provided in the question

Let us assume the mass of baseball = m = 0.145 kg

The Initial velocity of pitched ball = $v_i$ = 47 m/s

Final velocity of batted ball in the opposite direction = $v_f$ = -55m/s

Based on the above information, the change in momentum is

$\Delta P = m(v_f -v_i)$

$= 0.145 kg(-55m/s - 47m/s)$

= 14.79 kgm/s

Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s

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Answer:

Option (2) is correct.

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An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.

The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.

So, the correct option is (2).

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Please check the attached picture

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Answer:

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