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Marianna [84]
4 years ago
6

A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

3.5 rad/s. What is the magnitude of the angular acceleration of the wheel?
Physics
2 answers:
Mila [183]4 years ago
6 0

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

vfiekz [6]4 years ago
4 0

Use the formula

{\omega_f}^2-{\omega_i}^2=2\alpha\Delta\theta

\implies\left(13.5\dfrac{\rm rad}{\rm s}\right)^2-\left(22.0\dfrac{\rm rad}{\rm s}\right)^2=2\alpha(13.8\,\mathrm{rad})

\implies\alpha=-10.9\dfrac{\rm rad}{\mathrm s^2}

so the magnitude is 10.9 rad/s^2.

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A parallel-plate capacitor is charged by a 14.0 V battery, then the battery is removed. You may want to review (Pages 692 - 695)
Anika [276]

Answer:

Potential Difference = 14 V

Explanation:

We are told that when the capacitor plates are charged to a certain voltage, then we have;

ΔV = 14 volts

Now, the battery is disconnected, so here we have the potential difference between the plates to be given by the formula;

ΔV = Q/C

Now, the charge is conserved on the plates and the capacitance is constant, therefore in this case, the potential difference will remain the same.

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3 years ago
Contrast the atmosphere with the biosphere
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4 0
4 years ago
What does Δx stand for? What SI units are used to measure it?
pav-90 [236]

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4 0
3 years ago
A boat moving at 7.8 km/hr relative to the water is crossing a river 3.5 km wide in which the current is flowing at 2.8 km/hr. A
Free_Kalibri [48]

Answer:\theta =21.03^{\circ}

Explanation:

Given

velocity of boat with respect to river =7.8 km/hr

velocity of river is 2.8 km/hr

River is 3.5 km wide

Suppose boat leaves at an angle of \theta  with vertical such that it reaches exactly opposite end of initial Point

Therefore

7.8\sin \thetacomponent will balance the river Flowso that sin component helps to cross the river

7.8\sin \theta =2.8

\sin \theta =0.3589

\theta =sin^{-1}(0.3589)

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8 0
3 years ago
A 0.017-kg acorn falls from a position in an oak tree that is 18.5 meters above the ground. Calculate the velocity of the acorn
sdas [7]

Answer:

The velocity of the acorn just before it reaches the ground is 19 m/s

The kinetic energy when hitting the ground is 3.1 J

Explanation:

Given;

mass of the acorn, m = 0.017 kg

height of fall, h = 18.5 m

Apply the law of conservation of mechanical energy;

mgh = ¹/₂mv²

gh = ¹/₂v²

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Thus, the velocity of the acorn just before it reaches the ground is 19 m/s

Now, determine the kinetic energy when hitting the ground;

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K.E = ¹/₂(0.017)(19)²

K.E = 3.09 J

K.E = 3.1 J

Therefore, the kinetic energy when hitting the ground is 3.1 J

5 0
3 years ago
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