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stealth61 [152]
3 years ago
10

5 renewable energy resources i have nuclear, solar, wind what am i missin?

Physics
1 answer:
Tamiku [17]3 years ago
6 0

Answer:

Solar energy.

Wind energy.

Hydro energy.

Tidal energy.

Geothermal energy.

Biomass energy.

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Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237
Paladinen [302]

Answer:

Radius = 9.0216 cm

Explanation:

Given that:

The critical mass of neptunium-237 = 60 kg

Also, 1 kg = 1000 g

So mass = 60000 g

Density = 19.5 g/cm³

Volume = ?

So, volume:  

Volume=\frac {{Mass}}{Density}  

Volume=\frac {60000\ g}{19.5\ g/cm^3}  

The volume of the material = 3076.92308 cm³

The expression for the volume of the sphere is:

V=\frac {4}{3}\times \pi\times {(radius)}^3

3076.92308=\frac{4}{3}\times \frac{22}{7}\times {(radius)}^3

\frac{4}{3}\times \frac{22}{7}\times {(radius)}^3=3076.92308

88\times {(radius)}^3=64615.38468

{(radius)}=\sqrt[3]{\frac{64615.38468}{88}}

<u>Radius = 9.0216 cm</u>

4 0
4 years ago
The equation of motion for a simple harmonic oscillator (SHO) is:
N76 [4]

Answer:

Explanation:

ω = \sqrt{\frac{k}{m} }

k = 2.5 N/m

m = 10 kg

\omega = \sqrt{\frac{2.5}{10} }

ω = .5 rad /s

x(t) = A cos(ωt + φ₀)

When t = 0 , x(t) = 0

0 = A cos(ωx 0 + φ₀)

cos φ₀ = 0

φ₀ = π /2

x(t) = A cos(ωt +π /2 )

Putting the value of ω

x(t) = A cos(.5 t +π /2 )

Differentiating on both sides

dx(t)/dt = - .5 A sin(.5 t +π /2 )

v(t) = - .5 A sin(.5 t +π /2 )

Given t =0 , v(t) = -5 m/s

-5 = - .5 A sin(.5 x0 +π /2 )

-5 = - .5 A sinπ /2

A = 10 m

x(t) = 10 cos( .5 t +π /2 )

b )

when t = π ( 3.14 s )

x(t) = -  10 m

when t = 2π ( 6.28s )

x(t) = 0

when t = 3π ( 9.42 s )

x(t) =  10 m

and so on

8 0
4 years ago
5. A wave with peaks separated by .34 m has a wavelength of ________________m.
Eduardwww [97]

Answer: 0.17 I think

Explanation:

I asked a doctor

8 0
4 years ago
Fig. 28-9 shows the cross-section of a hollow cylinder of inner radius a = 5 cm and outer radius b = 7 cm. A uniform current den
Dmitrij [34]

Answer:

1.507×10⁻⁴ T

Explanation:

a = 5 cm = 0.05 m

b = 7 cm = 0.07 m

j = 1 A/cm²

Distance from magnetic field = d = 10 cm = 0.1 m

μ₀ = Vacuum permeability = 4π×10⁻⁷ H/m

Magnetic of hollow cylinder

\oint B.ds=\mu_0 I\\\Rightarrow B2\pi d=\mu_0 I\\\Rightarrow B2\pi d=\mu_0J\pi(b^2-a^2)\\\Rightarrow B=\frac{\mu_0J\pi(b^2-a^2)}{2\pi d}\\\Rightarrow B=\frac{4\pi\times 10^{-7}\times 1\times 10^{-4}\pi(0.07^2-0.05^2)}{2\pi 0.1}\\\Rightarrow B=1.507\times 10^{-4}\ T

∴ Magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder is 1.507×10⁻⁴ T

7 0
4 years ago
All protons in a vacuum have the same<br> 1) Speed<br> 2) Wavelength<br> 3) Energy<br> 4) Frequency
IRISSAK [1]
4). Frequency


I had a class on it.
5 0
3 years ago
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