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3 years ago

Within an internal combustion engine, the can-shaped component that moves up and down the cylinder

Physics

1 answer:

Mrrafil [7]3 years ago

6 0

This component is called the piston.

A piston is a cylindrical engine component that slides back and fourth in the cylinder bore by forces produced during the combustion process. The piston acts as a movable end of the combustion chamber.

Pistons are commonly made of a cast aluminum alloy for excellent and light weight thermal conductivity. Aluminum expands when heated and proper clearance must be provided to maintain free piston movement in the cylinder.

If the clearance is not sufficient the piston will seize and if it is excessive there will be loss of compression.

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A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so

joja [24]

ANSWER:100dB

from the sound intensity level,the sound intensity is calculated as:

$\beta$ ₁= $\beta$ =(10dB)㏒₁₀(l₁/l₀)

inserting numbers:

120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²

Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which

can be used to solve for l₁ and get

l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²

since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in

ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:

l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²

since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:

$\beta$ ₂= $\beta$ =(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)

$\beta$ ₂=(10dB)㏒₁₀(2.04×10¹⁰)

$\beta$ =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB

3 0

3 years ago

A charge of 4.5x10^-5 C is placed in an electric field with a strength of 2.0x10^4 N/C. What is the electric force acting on the

Alex

Answer:

0.9 N

Explanation:

The electric force acting on a charge is given by:

$F=qE$

where

q is the magnitude of the charge

E is the strength of the electric field

In this problem, we have

$q=4.5\cdot 10^{-5}C$ is the charge

$E=2.0\cdot 10^4 N/C$ is the strength of the electric field

Substituting into the equation, we find

$F=(4.5\cdot 10^{-5}C)(2.0\cdot 10^4 N/C)=0.9 N$

6 0

3 years ago

Problem 3) Bob stands at the edge of the swimming pool holding a laser 1.5m above the ground. He shines the red laser beam onto

Vadim26 [7]

Answer:

d = 5.75m

Explanation:

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

Here,

i = 90 - θ

$\theta = \tan^-^1(\frac{1.5}{3} )\\\\=26.56^\circ$

$r = \sin^{-1}\frac{n_1\sin i}{n_2}$

$r = \sin^{-1}\frac{(1)\sin (90-26.56)}{1.33}\\\\r = 42.26m$

$\tan r = \frac{2.5}{d_1}$

$d_1 = \frac{2.5}{\tan (42.26)} \\\\d_1 = 2.75m$

Therefore, the distance is

d = 3 + d₁

d = 3 + 2.75

d = 5.75m

6 0

3 years ago

Estimate the peak wavelength for radiation from ice at 273 k.

Andrews [41]

<h2>Answer: 10615 nm</h2>

Explanation:

This problem can be solved by the Wien's displacement law, which relates the wavelength $\lambda_{p}$ where the intensity of the radiation is maximum (also called peak wavelength) with the temperature $T$ of the black body.