How much tension must a rope withstand if it is used to accelerate a 960 kg car horizontally along a frictionless track at 1.20
2 answers:
Answer:
<em>The tension of the rope is 1152 N</em>
Explanation:
<u>Force</u>
Newton's second law states that the acceleration of an object is directly proportional to the net force and inversely related to its mass.
The law can be written as:
F = m.a
The rope must apply a force F to accelerate the car of m=960 Kg at . Given the track is frictionless, all the force applied produces acceleration, thus:
F = 1152 N
The tension of the rope is 1152 N
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Answer
given,
t = 3 s
we know,
position of the particle
integrating both side
Position of the particle at t= 3 s
x = 182.98 ft
Distance traveled by the particle in 3 s is equal to 182.98 ft
now, particle’s acceleration
at t= 3 s
a = 2.98 ft/s²
acceleration of the particle is equal to 2.98 ft/s²
Answer:
correct option is C. 1700 - 2300
Explanation:
solution
according to Care stream, exposure indicators provide the technical adequancy incident radiations on the x ray
and we muse care stream formula that is
Exposure Index EI = 1000 × + 2000
here 1 mR = 2000 EI
and 2 mR = 2300 EI
and 3 mR = 1700 EI
so here acceptable range of exposure indicators is 1700 - 2300
correct option is C. 1700 - 2300