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3 years ago

7

Suddenly she hears a pop from her tire. She gets off the bike and examines the tire - she notices a small hole in the tire and f

eels air coming from the hole.
The air flows out of the tire because the air pressure inside the tire is _______ the pressure outside the tire. Fill in the blank by choosing one answer below.

2 answers:

rewona [7]3 years ago

6 0

Answer:

b

Explanation:

vichka [17]3 years ago

3 0

I believe the answer is B. Have an awesome day!! Hope I helped you!

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Two girl scouts are sitting in a large canoe on a still lake while at summer camp. the canoe happens to be oriented with the fro

MrRissso [65]

That it's more puashe on the back of the canoe and that effects the back of the canoe to fall back

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3 years ago

What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?

ratelena [41]

Answer:

15.0 N

Explanation:

see pic

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3 years ago

A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k

Bad White [126]

Answer:

$v_{f,w} = 1.791\,\frac{m}{s}$ , $v_{f,c} = 0.972\,\frac{m}{s}$

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

$I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}$

$\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)$

$\omega_{f} \approx 5.116\,\frac{rad}{s}$

The tangential velocities of the wheel and the clay are, respectively:

$v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )$

$v_{f,w} = 1.791\,\frac{m}{s}$

$v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )$

$v_{f,c} = 0.972\,\frac{m}{s}$

5 0

3 years ago

A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 15

Sveta_85 [38]

Answer:

- the volume of the second tank is 1.77 m³

- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

Explanation:

Given that;

$V_{A}$ = 1 m³

$T_{A}$ = 10°C = 283 K

$P_{A}$ = 350 kPa

$m_{B}$ = 3 kg

$T_{B}$ = 35°C = 308 K

$P_{B}$ = 150 kPa

Now, lets apply the ideal gas equation;

$P_{B}$ $V_{B}$ = $m_{B}$ R $T_{B}$

$V_{B}$ = $m_{B}$ R $T_{B}$ / $P_{B}$

The gas constant of air R = 0.287 kPa⋅m³/kg⋅K

we substitute

$V_{B}$ = ( 3 × 0.287 × 308) / 150

$V_{B}$ = 265.188 / 150

$V_{B}$ = 1.77 m³

Therefore, the volume of the second tank is 1.77 m³

Also, $m_{A}$ = $P_{A}$ $V_{A}$ / R $T_{A}$ = (350 × 1)/(0.287 × 283) = 350 / 81.221

$m_{A}$ = 4.309 kg

Total mass, $m_{f}$ = $m_{A}$ + $m_{B}$ = 4.309 + 3 = 7.309 kg

Total volume $V_{f}$ = $V_{A}$ + $V_{B}$ = 1 + 1.77 = 2.77 m³

Now, from ideal gas equation;

$P_{f}$ = $m_{f}$ R $T_{f}$ / $V_{f}$

given that; final temperature $T_{f}$ = 20°C = 293 K

we substitute

$P_{f}$ = ( 7.309 × 0.287 × 293) / 2.77

$P_{f}$ = 614.6211119 / 2.77

$P_{f}$ = 221.88 kPa ≈ 222 kPa

Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa

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3 years ago

The number of wavelengths that pass a given point/second is called

myrzilka [38]

Answer:

I believe the answer is frequency

Hope this helps!! (let me know if it is right)

Explanation:

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3 years ago