Two car collide in an intersection. The speed limit in that zone is 30 mph. The car (mass of 1250 kg) was going 17.4 m/s (38.9).
1 answer:
Answer:
u₂ = 3.7 m/s
Explanation:
Here, we use the law of conservation of momentum, as follows:
where,
m₁ = mass of the car = 1250 kg
m₂ = mass of the truck = 2020 kg
u₁ = initial speed of the car before collision = 17.4 m/s
u₂ = initial speed of the tuck before collision = ?
v₁ = final speed of the car after collision = 6.7 m/s
v₂ = final speed of the truck after collision = 10.3 m/s
Therefore,
<u>u₂ = 3.7 m/s</u>
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A string with linear density 0.500 g/m.
Tension 20.0 N.
The maximum speed
The energy contained in a section of string 3.00 m long as a function of .
We are given following data for string with linear density held under tension :
μ = 0.5
= 0.5 x 10⁻³
T = 20 N
If string is L = 3m long, total energy as a function of is given by:
E = 1/2 x μ x L x ω² x A²
= 1/2 x μ x L x
= 7.5 x 10⁻⁴
So, The total energy as a function of = 7.5 x 10⁻⁴
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Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force
Mass of object = 2.00 kg
Initial position
Final position
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done
Put the value into the formula
Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.