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wlad13 [49]
3 years ago
6

A CD case slides along a floor in the positive direction of an x axis while an applied force a acts on it. The force is directed

along the x axis and has the x component Fax = 12x - 3x2, with x in meters and Fax in newtons. The case starts at rest at the position x = 0, and it moves until it is again at rest.
(A) Plot the work done on the case by a versus position x. (Do this on paper. Your instructor may ask you to turn in this work.)
(B) At what position is the work maximum?
(C) What is the maximum value of the work?
(D) At what position has the work decreased to zero?
(E) At what position is the case again at rest?
Physics
1 answer:
alisha [4.7K]3 years ago
4 0
Your answer should be b if I’m not mistaken
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I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT
Gala2k [10]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

5 0
3 years ago
the average power of the Sun is 3.79x1026 Watts, answer the following questions: 13. What is the average intensity of light at E
Lemur [1.5K]

Explanation:

Given that,

Average power of sun P=3.79\times10^{26}\ Watt

We need to calculate the intensity of light at Earth's position

Using formula of intensity

I=\dfrac{P_{avg}}{4\pi r^2}

Where, I = intensity

P = power

Put the value into the formula

I=\dfrac{3.79\times10^{26}}{4\pi\times(1.496\times10^{11})^2}

I=1347.616\ W/m^2

So, The intensity is 1347.616 W/m².

(A). We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully reflective

P = \dfrac{2I}{c}

Put the value into the formula

P=\dfrac{2\times1347.616}{3\times10^{8}}

P=8.984\times10^{-6}\ N/m

(B).  We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully absorptive

P=\dfrac{I}{c}

P=\dfrac{1347.616}{3\times10^{8}}

P=4.492\times10^{-6}\ N/m

Hence, This is the required solution.

6 0
3 years ago
What are the characteristics of an index fossil?
vichka [17]
I think as a mold. when the Flood came in Genesis, i believe that when the fish were washed away, the kinda made a mold in a rock.
4 0
3 years ago
Read 2 more answers
an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?
gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

F_{net} = F_g - F = m\cdot g - F\\

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N

The tension force F in the supporting cables is 7245N


3 0
4 years ago
A point charge q1=+5.00nC is at the fixed position x=0, y=0, z=0. You find that you must do 8.10×10−6J of work to bring a second
Maru [420]

The value of the second charge is 1.2 nC.

<h3>Electric potential</h3>

The work done in moving the charge from infinity to the given position is calculated as follows;

W = Eq₂

E = W/q₂

<h3>Magnitude of second charge</h3>

The magnitude of the second charge is determined by applying Coulomb's law.

E = \frac{kq_2}{r^2} \\\\\frac{kq_2}{r^2} = \frac{W}{q_2} \\\\kq_2^2 = Wr^2\\\\q_2^2 = \frac{Wr^2}{k} \\\\q_2 = \sqrt{\frac{Wr^2}{k} } \\\\q_2 =  \sqrt{\frac{(8.1 \times 10^{-6}) \times (0.04)^2}{9\times 10^9} } \\\\q_2 = 1.2 \times 10^{-9} \ C\\\\q_2 = 1.2 \ nC

Thus, the  value of the second charge is 1.2 nC.

Learn more about electric potential here: brainly.com/question/14306881

7 0
2 years ago
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