I'll bite:
-- Since the sled's mass is 'm', its weight is 'mg'.
-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is (μk) times (mg) .
-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.
-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)
-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.
-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write
s = (1/2 t²) (F - μk·mg) / m .
Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.
Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m
Power = (Work / time) = <em>F (t/2) (F - μk·mg) / m </em>
Unless I can come up with something a lot simpler, that's the answer.
To simplify and beautify, make the partial fractions out of the
2nd parentheses:
<em> F (t/2) (F/m - μk·m)</em>
I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.
Five points,ehhh ?
Answer:
d) It will be cut to a fourth of the original force.
Explanation:
The magnitude of the electrostatic force between the charged objects is
where
k is the Coulomb's constant
q1 and q2 are the charges of the two objects
r is the separation between the two objects
In this problem, the initial distance is doubled, so
r' = 2r
Therefore, the new electrostatic force will be
So, the force will be cut to 1/4 of the original value.
Answer:
A)0.00966 N/C
B) counterclockwise direction
Explanation:
We are given;
Diameter of the metal ring; d = 4.3 cm
Radius;r = 2.15 cm = 0.021- m
Initial magnetic field, B = 1.12 T
Rate of decrease of the magnetic field;dB/dt = 0.23 T/s
Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :
∫E•dr = d/dt∫B.A •dA
This gives;
E(2πr) = dB/dt(πr²)
Gives;. 2E = dB/dt(r)
E = dB/dt × 2r
We are given;
E = 0.23 × 2(0.021)
E = 0.00966 N/C
The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C
B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet
v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ;
and solve for "v".
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Explanation:
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The formula is: KE = (½) * (m) * (v²) ;
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"Kinetic energy" = (½) * (mass) * (velocity , "squared")
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Note: Velocity is similar to speed, in that velocity means "speed and direction"; however, if you "square" a negative number, you will get a "positive"; since: a "negative" multiplied by a "negative" equals a "positive".
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So, we have the formula:
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KE = (½) * (m) * (v²) ; to solve for "(v)" ; velocity, which is very similar to the "speed";
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we arrange the formula ;
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(KE) = (½) * (m) * (v²) ; ↔ (½)*(m)* (v²) = (KE) ;
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→ We have: (½)*(m)* (v²) = (KE) ; we isolate, "m" (mass) on one side of the equation:
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→ We divide each side of the equation by: "[(½)* (m)]" ;
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→ [ (½)*(m)*(v²) ] / [(½)* (m)] = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
to get:
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→ v² = (KE) / [(½)* (m)]
→ v² = 2 KE / m
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Take the "square root" of each side of the equation ;
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→ √ (v²) = √ { 2*(KE) ] / m }
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→ v = √ { 2*(KE) ] / m } ;
Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"];
and solve for "v".
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