Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
(a) Angular acceleration is 1.112 rad/s².
(b) Average angular velocity is 2.78 rad/s .
Explanation:
The equation of motion in Rotational kinematics is:
θ = θ₀ + 0.5αt²
Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.
(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:
13.9 = 0 + 0.5α(5)²
α = 1.112 rad/s²
(b) The equation of average angular velocity is:
ω = Δθ/Δt
ω = 
ω = 2.78 rad/s
Answer:
68.8 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of box = 18 Kg
Coefficient of friction (μ) = 0.39
Force of friction (F) =?
Next, we shall determine the normal force of the box. This is illustrated below:
Mass (m) of object = 18 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal force (N) =?
N = mg
N = 18 × 9.8
N = 176.4 N
Finally, we shall determine the force of friction experienced by the object. This is illustrated below:
Coefficient of friction (μ) = 0.39
Normal force (N) = 176.4 N
Force of friction (F) =?
F = μN
F = 0.39 × 176.4
F = 68.796 ≈ 68.8 N
Thus, the box experience a frictional force of 68.8 N.
Answer: Q=5.46 L/s
COP=2.58
Explanation:
Given that
Cp = 4.18 kJ/(kg.C
density = 1 kg/L
Heat rejected Qr= 570 kJ/min
Power in put W= 2.65 KW
From first law of thermodynamics
U = W+ q
q = Heat absorbed
U = internal energy
W = workdone
U = 570 kJ/min = 9.5 KW
9.5 = 2.65 + q
q = 6.85 KW
COP = q/W
COP = 6.58 / 2.65
COP=2.58
Lets take volume flow rate is Q
So mass flow rate of water m = ρ Q
q = m Cp ΔT
6.85 = 1 x Q x 4.18 ( 23-5)
Q=0.091 L/min
Q=5.46 L/s
Answer: large intestine
Explanation:
mouth grind and moisten food
saliva break down starch into sugar
esophagus transport food from mouth to stomach
stomach churn food, break down food with enzymes
liver add bile to break down fat
pancreas add enzymes to digest protein
gallbladder store bile
small intestine absorb food through villi
large intestine absorb extra water and remove waste from body
