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bonufazy [111]
3 years ago
11

The answer is 4 but idk how to get it so can somebody please explain how to get it

Physics
1 answer:
klemol [59]3 years ago
3 0
How much gravitational potential energy does the block have
when it gets to the top of the ramp ?

      (weight) x (height) = (15 N) x (0.2 m)  =  3 Joules .

If there were no friction, you would only need to do 3 Joules of work
to lift the block from the bottom to the top.

But the question says you actually have to do 4 Joules of work
to get the job done.

Friction stole one of your Joules along the way.

Choice-4 is not the correct one.
Choice-1 is the correct one.

===========================

Notice that the mass of the block is NOT  15 kg , and you
don't have to worry about gravity to answer this question.

The formula for potential energy is  (m)·(g)·(h) .
But (m·g) is just the WEIGHT, and the formula
is actually (weight)·(height).
The question GIVES us the weight of the block . . . 15 N .
So the potential energy at the top is just (15N)·(0.2m) = 3 Joules.
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A wave carries <u>energy</u><u> </u>from one place to another.

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Why is there more potential for wind energy in the upper himalayan in nepal?​
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An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, v_t = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

v_t = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, v_t = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

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