(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
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Answer:
Explanation:
We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .
rate of volume flow of one stream
= cross sectional area x velocity
= 8.4 x 3.5 x 2.2 = 64.68 m³ /s
rate of volume flow of other stream
= 6.6 x 3.6 x 2.7
= 64.15 m³ /s
rate of volume flow of rive , if d be its depth
= 11.2 x d x 2.8
= 31.36 d
volume flow of river = Total of volume flow rate of two streams
31.36 d = 64.15 + 64.68
31.36 d = 128.83
d = 4.10 m /s .
Answer:
8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.
Explanation:
The total charge accumulated in 1.8 hours will be:
Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)
Total Charge = - 12960 nC = - 12.96 x 10^(-6) C
Since, the charge on one electron is e = - 1.6 x 10^(-19) C
Therefore, no. of electrons will be:
No. of electrons = Total Charge/Charge on one electron
No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]
<u>No. of electrons = 8.1 x 10^13 electrons</u>
