the work done on the car by the force of its engine is 78,000 J.
" The work done on the car by the force of friction is 24,000 J.
Increasing the car's kinetic energy at the end of the first 100 m is 54,000J
a. Completed work = force x distance. Engine output = 780 N, that is,
780 N x 100 m = 78,000 J.
b. Completed work = force x distance. Friction force = 240 N, that is,
240 N x 100 m = 24,000 J.
c. Kinetic energy = 1 \ 2 x m x v2
= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.
<h3>How powerful is the engine of a car? </h3>
Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.
Learn more about work done here: brainly.com/question/25573309
#SPJ10
Answer:
A person climbs a set of stairs
Explanation:
Potential energy is said to be possessed by an object due to its position. As the height from the ground level increase, the potential energy increases. It is calculated by the below formula as :
P = mgh
Out of the given options, the option that illustrates an increase in potential energy is option (b) i.e. a person climbs a set of stairs. As he steps one stair, its position from ground increases. It means its potential energy increases.
How do you find instantaneous velocity
Select a point on a distance-time curve graph. Draw a tangent to the curve at that point. Tangent -> hypotenuse of right angled triangle. Opp/adjacent in graph units is vel at that point -> in distance and/or time
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.
The work will be completely defined by
(Force) x (distance in the y-direction),
and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.
We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.
From y=0 to y=2.40 is a distance of 2.40 upward.
Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.
Work = integral of (F·dy) evaluated from 0 to 2.40
= integral of (-2.85 y² dy) evaluated from 0 to 2.40
= (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .
Now, integral of (y² dy) = 1/3 y³ .
Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)
= 1/3 · 13.824 = 4.608 .
And the work = (-2.85) · the integral
= (-2.85) · (4.608)
= - 13.133 .
-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.
-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.
-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.
As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
Answer:
3.6km South East
Explanation:
Displacement is the shortest distance between the starting point and the ending point and the direction it is displaced in. To calculate the displacement we can use the pythagoras theorem because the 3km East and the 2km south form the two shorter sides of a right angled triangle between the starting and ending points. So, the displacement is the length C of the triangle which we can calculate as follows:
Pythagoras Theorem:
a^2+b^2=c^2
(2)^2+(3)^2=c^(2)
4+9=c^2
Square root 13 = c
c=3.6km (1dp)
The total displacement is 3.6km and is in the approximate direction of South East (because he travelled east and south).
Hope this helped!
