Answer:
3360 N
Explanation:
In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.
The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.
The torques are equal:
Fr = Fr
(1440 N) (3.5 m) = F (1.5 m)
F = 3360 N
Answer:
find V using d = M/V
Explanation:
F = d*V*g
d = density of fluid (in this case, 1000)
V = volume of object
g = gravity
ANSWER: Itna Bada answer Kisi Ko Pata chalega
EXPLANATION : please Manje brainliest karo man Jay he brainiest Karo .
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:
The magnetic field is 
Explanation:
From the question we are told that
The mass of the metal rod is 
The current on the rod is 
The distance of separation(equivalent to length of the rod ) is 
The coefficient of kinetic friction is 
The kinetic frictional force is 
The constant speed is 
Generally the magnetic force on the rod is mathematically represented as
For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so
=> 
=> 
=>
