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anygoal [31]
3 years ago
10

Roger pushes a box on a 30° incline. If he applies a force of 60 newtons parallel to the incline and displaces the box 10 meters

along the incline, how much work will he do on the box?
A.
5.2 × 102 joules
B.
6.0 × 102 joules
C.
-5.2 × 102 joules
D.
-6.0 × 102 joules
Physics
1 answer:
Arlecino [84]3 years ago
3 0

Answer:

The work done is, w = 6 x 10² J

Explanation:

Given,

The angle of the incline, ∅ = 30°

The force applied parallel to the incline, F = 60 N

The displacement caused by the force along the incline, s = 10 m

Since the force and displacement is along the incline, the angle of incline doesn't contribute to the work done.

The work done by the force is given by the relation,

                               W = F · S

                                    = 60 N x 10 m

                                    = 600 J

Hence, the work done by Roger pushing the box is, W = 600 J

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GuDViN [60]

Answer:

Assuming that you meant the final velocity of 50 m/s was reached in 10 s, the answer would be 5 m/s^2.

Explanation:

V_{f} = V_{i} + at

So we update that with the values that we have.

50 = 0 +a(10)

then simplify that using algebra to solve for a and we get 5 m/s^2

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A 1300 watt hair blow dyer is designed to operate on 120 Volts. How much current does the dryer require
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Answer:

10.83 Amperes

Explanation:

if   A ⇒ current

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3 years ago
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
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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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1) When you hold your nose and go underwater, you can still hear sounds that are made above the water, in the air, if they are l
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Answer:

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