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AlladinOne
1 month ago
10

A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s

and rolls in a straight line without slipping. If the moment of inertia of the coin is MR2 , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Physics
1 answer:
ValentinkaMS [3.4K]1 month ago
6 0
This question is incomplete. The query is regarding a 3.00 cm diameter coin rolling up a 30.0° incline. With an initial angular speed of 60.0 rad/s, it rolls without slipping. Given that the moment of inertia of the coin is (1/2) MR², the distance the coin travels up the incline is calculated as 0.124 m.
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Answer:

The convergence of light rays redirects them toward the focal point, resulting in a magnifying effect.

Explanation:

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2 months ago
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ????????,2=17.9vi,2=17.9
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2 months ago
A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

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

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0
2 months ago
Which statements describe properties of stars check all that apply
Keith_Richards [3271]

Answer:

Stars generate energy by the process of nuclear fusion.

They are large entities composed of gaseous elements.

The main constituents of stars are hydrogen and helium.

Explanation:

Stars are colossal objects with extensive gravitational forces causing them to contract, which allows fusion to take place: the atomic nuclei in the star's core are drawn very close together due to gravity and elevated temperatures, leading to the fusion reaction. This fusion serves as the energy output for a star.

Conversely, it is true that stars predominantly consist of hydrogen and helium (two hydrogen nuclei can fuse to become helium), which implies that a star is essentially an enormous ball of gas without a solid surface suitable for standing on.

As for the presence of water on a star, it is simply impossible. The extreme temperatures found in stars are far too high for water to exist in any liquid state on their surfaces.

7 0
2 months ago
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To find the mass using a weight of 1.4 N:
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3 0
3 months ago
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