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4 years ago

10

the gravitational force acting between two objects is 785 newton’s. if the mass of one object doubles, what is the new force act

ing between the objects

1 answer:

Tom [10]4 years ago

6 0

Use formula F=(G Mm)/R^2
from the relation When Mass of one object doubles the new gratification force will be 785*2=1570

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Answer:

5 m/s²

Explanation:

Acceleration = change in speed/time taken

20-5=15

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2 years ago

According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all ni

kodGreya [7K]

C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

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The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

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2 years ago

There is a bell at the top of a tower that is 45m high. The bell weighs 190N. The bell has ___________________ energy. Calculate

Arte-miy333 [17]

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3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

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$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

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F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

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$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

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$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

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3 years ago