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3 years ago

A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?

Physics

1 answer:

PSYCHO15rus [73]3 years ago

5 0

5x2.2/3 Because that's how you calculate velocity.

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How much heat (in calories) are needed to raise the temperature of 60 g of water

laila [671]

Answer:

Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius.

Explanation:

3 0

2 years ago

Productivity increases when A. inputs increase while outputs remain the same. B. inputs decrease while outputs remain the same.

Blizzard [7]

Answer: Productivity increases when inputs and outputs increase proportionately.

Explanation:

Productivity increases when inputs and outputs increase proportionately. Input has to be directly proportional to output to be productive. This means increase in input to a system must leads to drastic increase in the output. When the output is not balanced with the amount of input, it leads to unproductivity.

Being productive can be business wise or in terms if personal growth and development.

3 0

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.

- Then compute the distance from charges on the axis to the variable distance y:

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

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

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

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

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

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

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

- Now for the limit y >>a:

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

- Insert limit i.e a/y = 0

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

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

PLZ SOMEONEE HELPP I’LL MARK BRANLIESTTTT

7nadin3 [17]

Answer:

I'm pretty sure it's 37.5 joules of energy

Explanation:

hope this helps!

8 0

3 years ago

Which statement accurately describes impulse?<br> State corrrect ans

ser-zykov [4K]

Where are the statements at ?

5 0

3 years ago