Can you explain this a bit more I don’t quite understand
Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Answer:
I think the answer is D,54 joules
Answer:
Δ KE = 249158.6 kJ
Explanation:
given data
Truck mass M = 1560 Kg
Truck initial speed, u = 28 m/s
mass of car m = 1070 Kg
initial speed of car u1 = 0 m/s
solution
first we get here final speed by using conservation of momentum that is express as
Mu = (M+m) V .......................1
put here value we get
1560 × 28 = (1560 + 1070 ) V
solve it we get
final speed V = 16.60 m/s
and
Change in kinetic energy will be here
Δ KE =
.................2
put here value and we get
Δ KE =
solve it we get
Δ KE = 249158.6 kJ
Answer:
Machine Efficiency
Explanation:
Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work). The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent