What is the efficiency of a 0.60 kg basketball that, one dropped from a 2.0 m height (from basket rim) , rebounds to 1.2 m in he
1 answer:
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Answer:
Shortest time = 3.47 seconds
Explanation:
We know that Force of gravity of an object is given by;
F_g = mg
Also, Force due to kinetic friction is given as;
F_friction = μ_k•F_n
Force due to static friction is given as;
F_static = μ_s•F_n
Where ;
μ_k = coefficient of kinetic friction
μ_s = coefficient of static friction
F_n = Normal Force
Let's solve for the static crate.
From the free body diagram i attached, the resultant force in the vertical y direction will be;
N - mg = 0
Thus, N = mg
WherN is the
From earlier, we know that;
F_static = μ_s•F_n
Thus,
F_static = μ_s•mg
Now, in the horizontal x direction,
F_static = ma
Thus,
μ_s•mg = ma
m will cancel out, and;
a = μ_s•g
a = 0.3 x 9.81 = 2.943 m/s²
Not, to get the final the final speed of the tool box, we'll use;
V_f = V_i + a•Δt
Where;
Δt is shortest time.
V_f = final velocity
V_i = initial velocity
Thus, making Δt the subject, we have;
Δt = (V_f - V_i)/a = (10.2 - 0)/2.943 = 3.47 s
Answer:
(a). Check attachment.
(b). 280.305 J.
(c). 31.81 kpa; 38.26K.
(d). 24.05K.
(e). 24.05k; 40kpa.
(f). -138.6J.
Explanation:
(a). Kindly check the attached picture for the diagram showing the four process.
1 - 2 = adiabatic expansion process.
2 - 3 = Isochoric process.
3 - 4 = isothermal process.
4 - 1 = isochoric process.
(b). Recall that the process from 1 to is an adiabatic expansion process.
NB: b = 5/3 for a monoatomic gas.
Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].
= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.
Thus, the workdone = 280.305 J.
(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.
T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.
(d). The process 2 - 3 is an Isochoric process, then;
T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.
(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.
The pressure can be determine as below;
P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.
(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J
