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4 years ago

Please help answer the question in the screenshot. it's multiple choice..

Physics

1 answer:

Talja [164]4 years ago

7 0

Answer:

Since F = G * m1 * m2 / r^2

F = 6.67 * 19E-11 * 2.79 * 9.47 * 10E23 / (1.2 * 10^7)^2

F = 126 * 10E-2 N = 1.22 N

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Tension force ........... throughout a string that changes direction over a pulley.

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Answer:

"is constant"

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4 years ago

A blacksmith hammers a hot piece of iron while making a tool.How does the force due to hammering effect the piece of iron ???

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Answer:

The force due to hammering changes the shape of the piece of iron

Explanation:

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3 years ago

A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move

DochEvi [55]

Answer:

$W=-2.1405\times 10^9\,J$

Explanation:

Given:

electric field, $E=148\,V.m^{-1}$

charge, $Q=-13\,\mu C=-13\times 10^{-6}\,C$

initial position coordinates, $p1 =(-18,-131)$

final position coordinates, $p2 =(107,76)$

We find the distance through which the charge has been moved:

$d=\sqrt{(x1-x2)^2+(y1-y2)^2}$

Where we have (x1,y1) & (x2,y2) as the initial and final coordinates of the points.

$d=\sqrt{(107-(-81))^2+(76-(-131))^2}$

$d= 279.63\,m$

Now we need the angle through which displacement is made with respect to the direction of electric field.

$tan\,\theta= \frac{y2-y1}{x2-x1}$

$\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]$

$\theta= 47.75^{\circ}$

Now from the relation between the change in potential difference:

$\Delta V= E.d.cos\,\theta$

$\Delta V= 148\times 279.63\times cos\,47.75^{\circ}$

$\Delta V= 27826.06 V$

∵The change in voltage is defined as the work done per unit charge.

∴ $\Delta V=\frac{W}{Q}$

$W=\frac{\Delta V}{Q}$

Putting the respective values

$W=\frac{27826.06 }{-13\times 10^{-6}}$

$W=-2.1405\times 10^9\,J$

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3 years ago

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Answer:

Explanation:

1) No take-off possible

2) The downward force is greater than the upward force.

3) 2000 + 950 = 2950 N upward

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3 years ago

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3 years ago