Which describes the relationship between potential and kinetic energy of a ball thrown up in the air as it falls back to the gro
1 answer:
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Answer:
Explanation:
Given that,
Force is downward I.e negative y-axis
F = -2 × 10^-14 •j N
Magnetic field is westward, +x direction
B = 8.3 × 10^-2 •i T
Charge of an electron
q = 1.6 × 10^-19C
Velocity and it direction?
Force in a magnetic field is given as
F = q(V×B)
Angle between V and B is 270, check attachment
The cross product of velocity and magnetic field
F =qVB•Sin270
2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2
Then,
v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)
v = 1.51 × 10^6 m/s
Direction of the force
Let x be the direction of v
-F•j = v•x × B•i
From cross product
We know that
i×j = k, j×i = -k
j×k =i, k×j = -i
k×i = j, i×k = -j OR -k×i = -j
Comparing -k×i = -j to given problem
We notice that
-F•j = q ( -V•k × B×i)
So, the direction of V is negative z- direction
V = -1.51 × 10^6 •k m/s
Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,
Therefore, the work done by the worker in lifting the bucket is given as:
Now, plug in the values given and solve for 'W'. This gives,
Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.