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Contact [7]
3 years ago
15

You hike two thirds of the way to the top of a hill at a speed of 2.9 mi/h and run the final third at a speed of 5.6 mi/h. What

was your average speed?
Physics
1 answer:
Marta_Voda [28]3 years ago
3 0

Answer:

The average speed is 3.5 mi/h

Explanation:

Average speed is given by

Average speed = \frac{Total distance}{Total time}

If the total distance covered is x mi,

Then \frac{2}{3}x mi was covered while hiking and

\frac{1}{3}x mi was covered while running.

Now, we will find the time taken while hiking and the time taken while running

Speed = \frac{Distance}{ Time}\\  Time = \frac{Distance}{Speed}

  • For the time taken while hiking

Speed = 2.9 mi/h

Distance = \frac{2}{3}x mi

From,

Time = \frac{Distance}{Speed}

Time = \frac{\frac{2}{3}x }{2.9}

Time = 0.2299x h

Time taken while hiking is 0.2299 h

  • For the time taken while running

Speed = 5.6 mi/h

Distance = \frac{1}{3}x mi

Time = \frac{Distance}{Speed}

Time = \frac{\frac{1}{3}x }{5.6}

Time = 0.05952x h

Now, for the average speed

Average speed = \frac{Total distance}{Total time}

Total distance = \frac{2}{3}x mi  + \frac{1}{3}x mi = x mi

Total time = 0.2299x + 0.05952x = 0.28942x h

∴ Average speed = \frac{x}{0.28942x}

Average speed = 3.4552 mi/h

Average speed ≅ 3.5 mi/h

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Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

6 0
3 years ago
You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe
madam [21]

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

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3 years ago
A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A
Kruka [31]

Answer:

\omega_f = 585.37 \ rev/s

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)=  800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

I \omega_i = I' \omega_f

(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f

M\times 800 = ( M + m )\omega_f

3\times 800 = (3+1.1)\times \omega_f

2400 = (4.1)\times \omega_f

\omega_f = 585.37 \ rev/s

3 0
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Explain why the surface of Venus is hotter than the surface of Mercury
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<span>Venus is hotter due to the greenhouse effect: Venus has an atmosphere about ninety times thicker than that of Earth, and made almost entirely of carbon dioxide, which is one of the gasses that causes the greenhouse effect on Earth.</span>
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kolezko [41]

The rock strike the water with the speed of 15.78 m/sec.

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v=\sqrt{2gh}

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v=15.78 m/sec

Hence, the rock strike the water with the speed of 15.78 m/sec.

7 0
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