The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
<h3>What is the work done?</h3>
The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
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Answer:
Q1_new = 515.68 µC
Q2_new = 246.82 µC
Explanation:
Since the capacitors are charged in parallel and not in series, then both are at 250 V when they are disconnected from the battery.
Then it is only necessary to calculate the charge on each capacitor:
Q1 = 5.85 µF * 250 V = 1462.5 µC
Q2 = 2.8 µF * 250 V = 700 µC
Now, we will look at 1462.5 µC as excess negative charges on one plate, and 1462.5 µC as excess positive charges on the other plate. Now, we will use this same logic for the smaller capacitor.
When there is a connection of positive plate of C1 to the negative plate of C2, and also a connection of the negative plate of C1 to the positive plate of C2, some of these excess opposite charges will combine and cancel each other. The result is that of a net charge:
1462.5 µC - 700 µC = 762.5 µC
Thus,762.5 µC of net charge will remain in the 'new' positive and negative plates of the resulting capacitor system.
This 762.5 µC will be divided proportionately between the two capacitors.
Q1_new = 762.5 µC * (5.85/(5.85 + 2.8)) = 515.68 µC
Q2_new = 762.5 µC * (2.8/(5.85 + 2.8) = 246.82 µC
Answer:
2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)
Explanation:
The reaction is a neutralization reaction since an acid, aqeous H2SO4 reacts completely with an appropriate amount of alkali, aqueous KOH to produce salt, aqueous K2SO4 and liquid water, H2O only.
2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)
Alkali + Acid → Salt + Water.
During this reaction, 2 moles of KOH neutralize 1 mole of H2SO4 to yield 1 mole of K2SO4 and 2 moles of H2O.
Answer:
70509.8039216 N/C
Explanation:
k = Coulomb constant = 
q = Charge = 2.00 µC
l = Length of filament = 5.1 m
r = Radius of cylinder = 10 cm
Electric field is given by
The electric field at the surface of the cylinder is 70509.8039216 N/C
a) See free-body diagram in attachment
b) The acceleration is 
Explanation:
a)
The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.
The free-body diagram for this object is shown in the figure in attachment.
There are three forces acting on the object:
- The weight of the object, labelled as
(where m is the mass of the object and g is the acceleration of gravity), acting downward - The applied force,
, acting up along the plane - The force of friction,
, acting down along the plane
b)
In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:
where:
is the applied force, pushing forward
is the frictional force, acting backward
is the component of the weight parallel to the incline, acting backward, where
m = 2 kg is the mass of the object
is the acceleration of gravity
is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)
a is the acceleration
Solving for a, we find:
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