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Norma-Jean [14]
3 years ago
9

You throw a small toy ball that is covered with suction cups at the center of your glass patio door. When it hits it sticks to t

he glass, and because the door was not latched, it causes the door to swing open with an angular velocity of 0.19 rad/s. If the ball has a mass of 120 g and the patio door can be treated as a uniform box that is 2.3 m high, 1.0 m wide, and 0.030 m thick with a mass of 7.5 kg what speed did you throw the ball at
Physics
1 answer:
Natali [406]3 years ago
6 0

Answer:

Explanation:

moment of inertia of the door M = 1 /3 m ( l² + b² + d² )

= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )

= 1 / 3 x 7.5 x 6.2909

I = 15.7272.

moment of inertia of the door + small toy

15.7272 + .12 x .5²

= 15.7272 + .03

= 15.7572

Applying conservation of angular momentum law

mvr = M ω

m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.

.12 x v x .5 =  15.7572 x .19

v = 49.89 m /s .

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Answer

Hi,

In a chemical equation, chemicals that react are the reactants, while chemicals that are produced are the products/by products. Both sides of the equation must be balanced.

Explanation

When writing a chemical equation, reactants reacts to produce products. For example in the equation for formation of water, hydrogen combines with oxygen as 2H₂ +O₂→2H₂O where the first part before the arrow represent the reactants and the next part after the arrow are the products. Reactants are on the left where as products are on the right.Coefficient 2, in this cases is used for balancing the equation.

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4 0
3 years ago
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If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the
Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0
2 years ago
if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *
SSSSS [86.1K]

Answer:

0.013N

Explanation:

F=\frac{mv^{2} }{r}

m=0.04, v=4.4m/s, r=60

F=0.013 N

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3 years ago
Is it true or not true
maria [59]

Answer:

true

Explanation:

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3 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

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\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

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PART A)

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W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

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According to the value given we have that,

P = 4.0kW

P = 4000W

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P = \frac{W}{t}

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t = \frac{3004}{4000}

t = 0.75s

Therefore the interval for each cycle is 0.75s

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