This is not as simple as it looks.
What quantity are we going to compare between the two cases ?
Yes, I know ... the "amount of work". But how to find that from the
numbers given in the question ?
Is it the same as the change in speed ?
Well ? Is it ?
NO. IT's NOT.
In order to reduce the car's speed, the brakes have to absorb
the KINETIC ENERGY, and THAT changes in proportion to
the SQUARE of the speed. ( KE = 1/2 m V² )
Case 'A' :
The car initially has (1/2 m) (100²)
= (1/2m) x 10,000 units of KE.
It slows down to (1/2 m) x (70²)
= (1/2m) x 4,900 units of KE.
The brakes have absorbed (10,000 - 4,900) = 5,100 units of KE.
Case 'B' :
The car initially has (1/2 m) (79²)
= (1/2m) x 6,241 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 6,241 units of KE.
Just as we suspected when we first read the problem,
the brakes do more work in Case-B, bringing the car
to a stop from 79, than they do when slowing the car
from 100 to 70 .
But when we first read the problem and formed that
snap impression, we did it for the wrong reason.
Here, I'll demonstrate:
Change Case-B. Make it "from 70 km/h to a stop".
Here's the new change in kinetic energy for Case-B:
The car initially has (1/2 m) (70²)
= (1/2m) x 4,900 units of KE.
It slows down to a stop . . . NO kinetic energy.
The brakes have absorbed all 4,900 units of KE.
-- To slow from 100 to 70, the brakes absorbed 5,100 units of KE.
-- Then, to slow the whole rest of the way from 70 to a stop,
the brakes absorbed only 4,900 units of KE.
-- The brakes did more work to slow the car the first 30 km/hr
than to slow it the whole remaining 70 km/hr.
That's why you can't just say that the bigger change in speed
requires the greater amount of work.
______________________________________
It works exactly the same in the opposite direction, too.
It takes less energy from the engine to accelerate the car
from rest to 70 km/hr than it takes to accelerate it the
next 30, to 100 km/hr !
Answer:
2.21 N
Explanation:
The force in this case is the total mass multiplied by the acceleration due to gravity. You are not asked for the solution to be in terms of the torque which is the usual way to solve these problems. That's why you are not given where the fulcrum is.
The fulcrum feels F1 + F2 + 34 * 980
F2 = 141.7 * 980 = 138866
F1 = 50.3 * 980 = 49294
Ruler = 34 * 980= 33320
Total Force = 221480 The units here are dynes
I just saw in the middle of the question that g = 9.80
So the answer becomes 221480 / 1000 = 221.48 because we needed kg
And that answer becomes 221.48/100 2.21 because the force of gravity should be 9.8 not 980
The total force exerted on the fulcrum is
Answer:
The expression is Vmem = ΔV - I*Rax
Explanation:
According to the picture, if switch S is closed, the Cmem will be short. If Vcap = 0, the current flows through the capacitor only (not through Rmem), thus Vmem = 0 after closing S.
When C is fully charged, then we have:
Vmem = ΔV - I*Rax
Answer:
3.7 km
62.7175453187 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
The maximum altitude of the rocket is 750+875+2038.73598369 = 3663.73598369 m = 3.7 km
Time taken to reach the ground from max height 27.3301854818 seconds
Time to reach max height during free fall after stage 2 ends is 20.3873598369 s
Total time of flight is 10+5+20.3873598369+27.3301854818 = 62.7175453187 seconds
The formula for the volume of a cylinder is V=Bh or V=πr2h . The radius of the cylinder is 8 cm and the height is 15 cm. Substitute 8 for r and 15 for h in the formula V=πr2h . Simplify.