If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.
If you apply 1 newton to the machine's input, the output force is
(2.5 x 1 newton) = 2.5 newtons.
If you apply 120 newtons to the machine's input, the output force is
(2.5 x 120 newtons) = 300 newtons.
Answer:
Explanation:
Given two mass on an incline code 
and 
and an angle of inclination 
. 
. Assume that is the weight being pulled up and the hanging weight.
-The equations of motion from Newton's Second Law are:
where a is the acceleration.
#Substituting for 
(tension) gives:
#and solving for 
which is the system's acceleration.
-- The car starts from rest, and goes 8 m/s faster every second.
-- After 30 seconds, it's going (30 x 8) = 240 m/s.
-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s
-- Distance covered in 30 sec at an average speed of 120 m/s
= <span> 3,600 meters .</span>
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The formula that has all of this in it is the formula for
distance covered when accelerating from rest:
Distance = (1/2) · (acceleration) · (time)²
= (1/2) · (8 m/s²) · (30 sec)²
= (4 m/s²) · (900 sec²)
= 3600 meters.
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When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.
When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.
How does he do that ?
By accelerating at 8 m/s². That's about 0.82 G !
He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !
He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
Answer:
A) d_o = 20.7 cm
B) h_i = 1.014 m
Explanation:
A) To solve this, we will use the lens equation formula;
1/f = 1/d_o + 1/d_i
Where;
f is focal Length = 20 cm = 0.2
d_o is object distance
d_i is image distance = 6m
1/0.2 = 1/d_o + 1/6
1/d_o = 1/0.2 - 1/6
1/d_o = 4.8333
d_o = 1/4.8333
d_o = 0.207 m
d_o = 20.7 cm
B) to solve this, we will use the magnification equation;
M = h_i/h_o = d_i/d_o
Where;
h_o = 3.5 cm = 0.035 m
d_i = 6 m
d_o = 20.7 cm = 0.207 m
Thus;
h_i = (6/0.207) × 0.035
h_i = 1.014 m
The efficiency of an ideal Carnot heat engine can be written as:
where
is the temperature of the cold region
is the temperature of the hot region
For the engine in our problem, we have
and
, so the efficiency is
