An object with a mass of 30 kg moves at a rate of 15 m/S over distance of 10 m. 900 joules of work was done in 10 seconds. What
1 answer:
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Answer:
(a): the mug hits the floor 0.752m away from the end of the bar.
(b): the speed of the mug at impact are:
V= 4.87 m/s
direction= 70.82º below the horizontal.
Explanation:
Vx= 1.6 m/s
Vy=?
h= 1.1 m
g= 9.8 m/s²
t is the fall time
t=0.47 sec
Vy= g*t
Vy= 4.6 m/s
V= 4.87 m/s
α= tan⁻¹(Vy/Vx)
α= -70.82º
Answer:
(a). The work done is 7001 MeV.
(b). The momentum of this proton is .
Explanation:
Given that,
Speed = 0.993 c
We need to calculate the work done
Using work energy theorem
The work done is equal to the kinetic energy relative to the proton
Put the value into the formula
(b). We need to calculate the momentum of this proton
Using formula of momentum
Put the value into the formula
Hence, (a). The work done is 7001 MeV.
(b). The momentum of this proton is .
Answer:
Explanation:
Given
magnitude of vector=8.7 m/s
Direction
Therefore vector x component is
=-2.104
i.e. it is acting in negative x axis
vector y component
Therefore it lies in negative y axis
Judging vector x and y component
It lies in third quadrant