Answer:
Explanation:
First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:
Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.
Since 
and 
, we can rewrite the first equation as:
Now, we solve for 
and calculate it:
This means that the crate's coefficient of kinetic friction on the floor is 0.18.
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:
Torques must balance
F1 * X1 = F2 * Y2
or M1 g X1 = M2 g X2
X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7
X2 = 22.4 cm
Torque = F1 * X2 =
62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2
Normally x cross y will be out of the page
r X F for F1 will be into the page so the torque must be negative
A, D is the correct answers
