Block A with mass mA=4.6 kg lies on a frictionless incline plane. It is connected by a massless rope over a frictionless, massle
ss pulley to block B with mass mB=6.4 kg. The plane is inclined at an angle of θ = 53.9°. The blocks are released from rest and block B falls a distance L=2.9 m. Find the speed of block B (v) when it has fallen a distance L, and the total kinetic energy (K) of the two blocks at that instant.
1 answer:
Answer:
Check the explanation
Explanation:
In calculate kinetic energy, write out a formula whereby the kinetic energy will be equal to 0.5 times mass times velocity squared. include in the value for the object mass, then the velocity with which it is moving.
Kindly check the attached image below to get the step by step explanation to the question above.
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Answer:
a) 25.5°(south of east)
b) 119 s
c) 238 m
Explanation:
solution:
we have river speed =2 m/s
velocity of motorboat relative to water is =4.2 m/s
so speed will be:
a) =
+
solving graphically
=4.7 m/s
Ф=
=25.5°(south of east)
b) time to cross the river: t==
=119 s
c) d==(2)(119)=238 m
note :
pic is attached
Answer:
6.04788 in
78.38779 m/s
0.88159 kg
34.55294 J
Explanation:
Circumference is given by
Diameter is given by
The diameter is 6.04788 in
Volume of sphere is given by
The volume is
Fall velocity is given by
The velocity of the fall will be 78.38779 m/s
Mass is given by
The mass is 0.88159 kg
Kinetic energy is given by
The kinetic energy is 34.55294 J