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Olegator [25]
2 years ago
13

A robin flies a distance of 45963 cm. How far has it flown in kilometers?

Physics
1 answer:
alina1380 [7]2 years ago
6 0

Answer:

0.46km

Explanation:

45963cm/100cm=459.63m/1000m=0.45963 or 0.46km

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A substance changes from liquid to gas
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Explain briefly why metals are good conductors of electric current
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The free electrons in metals can move through the metal, all while receiving and losing electrons, allowing metals to conduct electricity. Example: copper is a great conductor of electric current.
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A machine accelerates a 5 kg missile from rest to a speed of 5 km/s. The net force accelerating the missile 500,000 N. How long
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Final velocity = 5km/sec = 5000 m/sec
Initial velocity = 0, mass = 5kg and force = 500,000 N
F = m(Vf-Vi)/time
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So, it takes 0.05 sec to arrive at the speed of 5km/s.
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What is the amount of charge a capacitor can store per volt of potential difference?
sladkih [1.3K]

The amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

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Hence, the amount of charge a capacitor can store per volt of potential difference is called the capacitance of the capacitor.

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2 years ago
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For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate
faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue=h_1=0.200 m

Metal volume flow  rate,Q=0.03m^3/min

Q=\frac{0.03}{60}=5\times 10^{-4}m^3/s because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

d_=0.030 m

h_2=0

A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}

A_1=7.065\times 10^{-4} m^2

v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}

v_1=0.708 m/s

Pressure at the top and bottom of the sprue is atmospheric

h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}

Substitute the values

0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}

v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264

v_2=\sqrt{4.421264}=2.1 m/s

Q=A_2v_2

5\times 10^{-4}=A_2\times 2.1

A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2

Reynolds number=\frac{v_2D\rho}{\eta}

\eta=0.004 N.s/m^2

\rho=2700 kg/m^3

Substitute the values then we get

Reynolds number=\frac{2.1\times 0.03\times 2700}{0.004}

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0
3 years ago
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