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xxMikexx [17]
2 years ago
13

The Sankey diagrams below show the energy transfers in two light bulbs. How much energy is wasted by light bulb 2?

Physics
1 answer:
Law Incorporation [45]2 years ago
5 0

Answer:

The amount of energy wasted by light bulb 2 = 70 J

Explanation:

Sankey diagrams are used to graphically indicate the proportional flow rate of a quantity

The labelled Sankey diagrams, here, indicate the energy transfers in two light bulbs, light bulb 1 and light bulb 2

From the labelled Sankey diagrams, we have;

For light bulb 1, the input energy = 80 J

The useful energy = 40 J

The wasted energy = 80 - 40 = 40 J

For  light bulb 2, the input energy = 80 J

The useful energy = 10 J

Therefore, the wasted energy = 80 - 10 = 70 J

The amount of energy wasted by light bulb 2 = 70 J.

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The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the
Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

      F = m a = m dv / dt

      dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

       dv = ∫ (3 + 2t) dt / m

       v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit   t = t and v = 0

       0 - (-6) = (3 (t- 0) + (t² -0)) / m

       t² + 3t -6m = 0

Let's look for the mass

      W = mg

      m = W / g

      m = 20/32

      m = 0.625 slug

Let's solve the second degree equation

     t² + 3t -3.75 = 0

     t = (-3 ± √ (32 + 4 1 3.75)) / 2

     t = (-3 ± 4,899) / 2

     t₁ = 0.95 s

     t₂ = -3.95 s

We take the positive time

6 0
2 years ago
A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a
Contact [7]

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

5 0
2 years ago
A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is mov
zzz [600]

Answer:

0.22m/s

Explanation:

The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.

momentum of an object = mass* velocity

Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;

Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;

Solve for x.

4 0
2 years ago
Aldis is swinging a ball tied to the end of a string over his head. Suddenly, the string breaks and the ball flies away. Arrow b
prisoha [69]

Answer:

Straight line in the direction of the tangential velocity the ball had at the moment the string broke

Explanation:

After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.

3 0
2 years ago
Read 2 more answers
Imagine you derive the following expression by analyzing the physics of a particular system: a=gsinθ−μkgcosθ, where g=9.80meter/
BARSIC [14]

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

8 0
2 years ago
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