A 5.22-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.10 m/s and its y component o
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Answer:
The magnitude of the tension on the ends of the clothesline is 41.85 N.
Explanation:
Given that,
Poles = 2
Distance = 16 m
Mass = 3 kg
Sags distance = 3 m
We need to calculate the angle made with vertical by mass
Using formula of angle
We need to calculate the magnitude of the tension on the ends of the clothesline
Using formula of tension
Put the value into the formula
Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.
Refer to the diagram shown below.
The net force acting on the vehicle is
F - R = 1060 -1010 = 50 N
The distance traveled is 21 m. Because the force is constant, the work done is
W = (50 N)*(21 m) = 1050 J
Assume that energy is not dissipated by air resistance or otherwise.
Conservation of energy requires that W = KE, where KE is the kinetic energy of the vehicle.
The KE is
KE = (1/2)*(2000 kg)*(v m/s)² = 1000v² J
Equate KE and W to obtain
1000v² = 1050
v² = 1.05
v = 1.025 m/s
Answer: 1.025 m/s
The net force acting on the vehicle is
F - R = 1060 -1010 = 50 N
The distance traveled is 21 m. Because the force is constant, the work done is
W = (50 N)*(21 m) = 1050 J
Assume that energy is not dissipated by air resistance or otherwise.
Conservation of energy requires that W = KE, where KE is the kinetic energy of the vehicle.
The KE is
KE = (1/2)*(2000 kg)*(v m/s)² = 1000v² J
Equate KE and W to obtain
1000v² = 1050
v² = 1.05
v = 1.025 m/s
Answer: 1.025 m/s