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kifflom [539]
3 years ago
7

How much time is needed for a 15,000 W engine to do 1,800,000 J of work? (Power: P = W/t)

Physics
2 answers:
vovikov84 [41]3 years ago
4 0
To solve for this we need to use the formula P = W/T (Power = Work/Time).
Since we do not have time, we shall switch up the formula.
Our new formula is T = W/P (Time = Work/Power).
We have 1,800,000 J of work, and 15,000 W of power.
1,800,000/15,000 = 120.
It will take 120 (insert measure of time here).
I hope this helps!
sergiy2304 [10]3 years ago
4 0

your answer would be 120

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*Look at attachment for photo of object**
prohojiy [21]

A. The magnitude of the spring force (in N) acting upon the object is 15.9 N

B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²

C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).

<h3>A. How to determine the force </h3>
  • Extension (e) = 0.150 m
  • Spring constant (K) = 106 N/m
  • Force (F) = ?

F = Ke

F = 106 × 0.15

F = 15.9 N

<h3>B. How to determine the acceleration</h3>
  • Mass (m) = 0.52 Kg
  • Force (F) = 15. 9 N
  • Acceleration (a) =?

F = ma

Divide both sides by m

a = F / m

a = 15.9 / 0.52

a = 30.58 m/s²

<h3>C. How to determine the direction of the acceleration vector</h3>

Considering the diagram, we can see that the spring was pulled away from the equilibrium point.

Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.

Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.

Learn more about spring constant:

brainly.com/question/9199238

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2 years ago
A rock is thrown upward with a velocity of 18 meters per second from the top of a 38 meter high cliff, and it misses the cliff o
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The rock would be at a point 12 m from water at a time <u>4.8 s</u>.

Take the origin of the coordinate system at the top of the cliff. It is thrown upwards with a velocity u. When the rock is at a point 12 m from water, calculate the vertical displacement of the rock from the origin.

y= -38m- (-12 m)=-26 m

Use the equation of motion,

y=ut +\frac{1}{2} at^2

The rock falls under the acceleration due to gravity, directed down wards.

Substitute 18 m/s for u, -26 m for y and -9.8 m/s² for a=g.

y=ut +\frac{1}{2} at^2\\ (-26 m)=(18m/s)t+\frac{1}{2}(-9.8m/s^2)t^2

Solve the quadratic equation for t.

t=\frac{(18m/s)(+/-)\sqrt{(18m/s)^2-4(4.9m/s^2)(-26m)} }{2(4.9m/s^2)}

Taking only the positive value,

t=\frac{(18 m/s)+(28.87 m/s)}{9.8 m/s^2)} \\ t=4.78 s

After a time of <u>4.8 s</u> the rock would be at a distance of 12 m from water.

8 0
3 years ago
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Answer:

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3. Static electricity is produced by charges(at rest) Magnetism is produced by charges(moving with uniform speed)

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