Given conditions:
height of object = 7.5cmdistance of object from mirror = 14 cmfocus length = -7 cmimage distance = ?
Using mirror formula:
1/(focus length) = 1/(object distance) + 1/(image distance)
or, -1/7 = 1/14 + 1/(image distance)
or, image distance = -4.66cm (the image formed is a virtual image)
Also, magnification of image is:
image height /height of object = - image distance /object distance
or, image height = - image distance / object distance * height of object
or, image height = -(-4.66) / 14 * 7.5 = 2.49 = 3(nearest whole number)
Vertical Free Fall and Constant Horizontal Motion
Density = (mass) / (volume) <== MEMORIZE THIS !
1). Mass = 50 g. Volume = 100 cm³. Density = (mass) / (volume)
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2). Volume = (length) ·(width) ·(height) = (4cm) ·(4cm) ·(4cm) = 64 cm³
Mass = 672 g. Density = (mass) / (volume)
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3). Volume = (length) ·(width) ·(height)
Length = 1 meter = 100 cm
Width = 10 cm = 10 cm
Height = 22 mm = 2.2 cm
Volume = (100 cm) (10 cm) (2.2 cm) = 2,200 cm³
Mass = 42,460 g
Density = (mass) / (volume)
(a) 154.5 N
Let's divide the motion of the sprinter in two parts:
- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration 
for a total time 
During this part of the motion, he covers a distance equal to 
, until he finally reaches a velocity of 
. We can use the following suvat equation:
which reduces to
(1)
since u = 0.
- In the second part, he continues with constant speed 
, covering a distance of 
in a time 
. This part of the motion is a uniform motion, so we can use the equation
(2)
We also know that the total time is 10.0 s, so
Therefore substituting into the 2nd equation
From eq.(1) we find
(3)
And substituting into (2)
Solving for t,
So from (3) we find the acceleration in the first phase:
And so the average force exerted on the sprinter is
b) 14.5 m/s
The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation
where we have
u = 0
is the acceleration
is the time of the first part
Solving the equation,
As you crank the variable resistor to higher resistance, the total current in the loop decreases. The power dissipated by the light bulb ... the heat and brightness it produces ... depend directly on the current through it, so they decrease too. Your circuit is a perfect incandescent light dimmer circuit.
