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Alexandra [31]
3 years ago
7

A particular wire has a resistivity of 6.47×10-8 Ωm and a cross-sectional area of 2.32 mm2. A length of this wire is to be used

as a resistor that will develop 130 W of power when connected to a 9.00 V battery. What length of wire is required?
Physics
1 answer:
PolarNik [594]3 years ago
6 0

Using  the formula for power:

P = V^2 / R  

130 W = (9.00 V)^2 / R  

Solve for r:

R = 81/130

R = 0.623 ohms

Now solve for the length of wire:

R = rho L / A  

A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2  

Now you have:

0.623 = (6.47x10^-8) L / (2.32x10^-6)  

L = 22.34 m (Round answer as needed.)

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Explanation:

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Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth,
Alex17521 [72]

Answer:

A) 1.67 x 10 ⁻⁶ m/s

B)5.59 x 10^-^9 %

Explanation:

A)

Given:

d = 5.0 km,

mₐ = 2.5 x 10^1^4 kg

u₁ = 4.0 x 10⁴ m/s

m_n = 5.98 x 10 ²⁴ kg

Solve using kinetic conserved energy

mₐ x u₁ + m_n  x u₂ = uₓ x (mₐ + m_n )

(2.5 x 10^1^4) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x 10^1^4 + 5.98 x 10 ²⁴ )

uₓ = ( 2.5 x 10^1^4 x 4.0 x 10⁴ ) / (2.5 x 10^1^4 + 5.98 x 10 ²⁴ )    

uₓ = 1.67 x 10 ⁻⁶ m/s

B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m

t = 365 days  x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s

t = 31536000 s

D = 2 π R = 2 π( 1.5 x 10 ¹¹ )

D = 9.4247 x 10 ¹¹ m

u₂  = D / t  = 9.4247 x 10 ¹¹  / 31536000

u₂ =  29885.775 m/s

% = (  1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100

% = 5.59 x 10^-^9 %

5 0
3 years ago
Help me with this please
pochemuha

Answer:

1. 31.5

3. 3.5

Explanation:

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