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xenn [34]
3 years ago
14

Which energy source has no greenhouse gas emissions or health impacts on humans, but is only

Physics
2 answers:
NARA [144]3 years ago
7 0

Answer:

i mean i would think solar engery example solar panels

Explanation:

man i tried for you sorry if i wasnt much help

yaroslaw [1]3 years ago
4 0

Answer: geothermal

Explanation:

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A 117 kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 62.
Ivenika [448]

Answer:

I_syst = 278.41477 kg.m²

Explanation:

Mass of platform; m1 = 117 kg

Radius; r = 1.61 m

Moment of inertia here is;

I1 = m1•r²/2

I1 = 117 × 1.61²/2

I1 = 151.63785 kg.m²

Mass of person; m2 = 62.5 kg

Distance of person from centre; r = 1.05 m

Moment of inertia here is;

I2 = m2•r²

I2 = 62.5 × 1.05²

I2 = 68.90625 kg.m²

Mass of dog; m3 = 28.3 kg

Distance of Dog from centre; r = 1.43 m

I3 = 28.3 × 1.43²

I3 = 57.87067 kg.m²

Thus,moment of inertia of the system;

I_syst = I1 + I2 + I3

I_syst = 151.63785 + 68.90625 + 57.87067

I_syst = 278.41477 kg.m²

8 0
3 years ago
A gas occupies a volume of 20 cubic meters at 9,000 pascals. If the pressure is lowered to 5,000 pascals, what volume will the g
forsale [732]
We need to consider no change in the temperature of gas (isothermal transformation)

Volume and pressure are inversely proportional magnitudes, so we can write:

P_1.V_1=P_2.V_2\\
\\
9.20=5.V_2\\
\\
V_2=\frac{180}{5}=36 \ m^3
5 0
2 years ago
A coin with a diameter 3.00 cm rolls up a 30.0 inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s
sweet [91]

This question is in complete.The question is

A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.

Answer:

distance=0.124 m

Explanation:

mgh=mglSin\alpha =(1/2)Iw_{i}^{2}+(1/2)mv^{2}\\   v=wR\\Solve for L\\L=((1/2)(1/2)0.015^{2}*60^{2}+(1/2)(60*0.015^{2} ))/9.8Sin30\\   L=0.124m

6 0
3 years ago
Numerical Problems:
dangina [55]
  • Displacement=1200m
  • Time=4min=4(60)=240s

\boxed{\sf Velocity=\dfrac{Displacement}{Time}}

\\ \sf\longmapsto Velocity=\dfrac{1200}{240}

\\ \sf\longmapsto Velocity=5m/s

6 0
3 years ago
In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ????
34kurt

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

I = \frac{q}{t}

<u>Where</u>:

q: is the electric charge transferred through the surface

t: is the time      

The charge, q, is:

q = n*e

<u>Where</u>:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C

Hence, the current in the wire is:

I = \frac{126.88 C}{3.97 s} = 31.96 A

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

3 0
2 years ago
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