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Lina20 [59]
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?
Physics
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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Just look it up for this page and maybe you will find an anwser sheet.

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1. Al descomponerse su vehículo una persona tira de su auto con la ayuda de una cuerda con una fuerza de 3500 N que forma un áng
OleMash [197]

Responder:

Fy = 2474,8737

Fx = 2474,8737

Explicación:

Dado que :

Dado:

Fuerza, F = 3500 N

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El componente vertical Fy; y

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8 0
3 years ago
Stairway must have uniform riser height and tread depth; variations in riser height or tread depth shall not be over _______ inc
alexandr1967 [171]

Answer:

\frac{3}{8} inches

Explanation:

the variations in riser height or tread depth should not be grater than \frac{3}{8} inches that is equal to 9.5 mm but the maximum riser height should be the  \frac{81}{4} inch  but variation in riser height should not exceed to \frac{3}{8} inches. The minimum riser height should be 7 inches which is equal to the 178 mm

5 0
4 years ago
A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats
Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

\lambda = \frac{c}{f}

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m            

Hence, the distance is:

d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

 

I hope it helps you!

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