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3 years ago

8

A gas-powered lawnmower is rated at 98 db. how many times more intense is the sound of this mower than that of an electric-power

ed mower rated at 70 db?

1 answer:

givi [52]3 years ago

6 0

The gas- powered mower is

(98dB - 70dB) = 28 dB louder.

28 dB = 10^2.8

= 631 times as much sound power.

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Explanation:

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3 years ago

A pitcher throws a curveball that reaches the catcher in 0.61 s. The ball curves because it is spinning at an average angular ve

sashaice [31]

Answer:

22.36 rad

Explanation:

Applying,

ω = θ/t.............. Equation 1

Where ω = angular velocity, θ = angular displacement of the baseball, t = time

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θ = ωt............... Equation 2

From the question,

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4 0

3 years ago

What is the resistance of : A) A 1.70 m long copper wire that is 0.700 mm in diameter? B) A 20.0 cm long piece of carbon with a

astra-53 [7]

Answer:

(I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

Explanation:

Given that,

Length of copper wire = 1.70 m

Diameter = 0.700 mm

Length of carbon piece = 20.0 cm

Cross section area $A = (2.00\times10^{-3})^2\ m$

(I). We need to calculate the area of copper wire

Using formula of area

$A=\pi r^2$

$A=3.14\times(\dfrac{0.700\times10^{-3}}{2})^2$

We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{1.68\times10^{-8}\times1.70}{3.14\times(\dfrac{0.700\times10^{-3}}{2})^2}$

$R=0.0742\ \Omega$

(II). We need to calculate the resistance

Using formula of resistance

$R=\dfrac{\rho l}{A}$

Put the value into the formula

$R=\dfrac{3.5\times10^{-5}\times20\times10^{-2}}{(2.00\times10^{-3})^2}$

$R=1.75\ \Omega$

Hence, (I). The resistance of the copper wire is 0.0742 Ω.

(II). The resistance of the carbon piece is 1.75 Ω.

8 0

3 years ago