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Mandarinka [93]
3 years ago
14

The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a

velocity of 3.6 m/s in the positive x direction, and some time later has a velocity of 7.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time

Physics
2 answers:
kumpel [21]3 years ago
5 0

Explanation:

Below is an attachment containing the solution.

bekas [8.4K]3 years ago
3 0

Answer:

The work done on the canister by the 5.0 N force during this time is

54.06 Joules.

Explanation:

Let the initial kinetic energy of the canister be

KE₁ = \frac{1}{2} mv_1^{2} = \frac{1}{2} *3*3.6^{2} = 19.44 J in the x direction

Let the the final kinetic energy of the canister be

KE₂ = \frac{1}{2} mv_2^{2} = \frac{1}{2} *3*7.0^{2} = 73.5 J in the y direction

Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final

= 73.5 J - 19.44 J = 54.06 J

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P= 10m = 10 x 9.10 = 91 N

Work is done lifting a 9.10-kg box straight up onto a shelf that is 1.80 m high :

A= Ph = 91 x 1.80 = 163.8 J

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5 0
2 years ago
A gate at the bottom of a dam is 10 ft tall by 30 ft wide. The specific weight of the water is 62.4 lbf/ft3. If the water is 50
Burka [1]

Answer:

The total force is most nearly 940,000lbf

Explanation:

Total force = specific weight × volume

Volume = length × width × depth = 10ft × 30ft × 50ft = 15,000ft^3

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4 0
3 years ago
At the bottom of a large cylindrical tank filled with fresh water the gauge pressure is 11.6 psi. What is the height (in feet) o
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Answer:

8.19m

Explanation:

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3 0
2 years ago
Read 2 more answers
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