Answer:
0.0192A
Explanation:
Since, the reading of the galvanometer is 0 A, the voltage across the resistance R will be:
Step 1
VR = V2
VR = 3.00v
Step 2
Calculating the current through the resistance R as below,
IR = V1 - V2 /R1
IR = 12 - 3 /468
IR =0.0192A
Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
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The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .
If an object starts from rest and its speed after 2 seconds is 20 m/s, then its acceleration is 20/2 = 10 m/s^2 .
(Gravity on Earth is only 9.8 m/s^2.)
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '.
D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens
to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .
-- They don't change by the same factor, because 1/g is inside
the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value
of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds
to roll off the same window sill and fall 120 meters down to the
surface of the Moon.
