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iren2701 [21]
3 years ago
10

Which type of mechanical wave moves back and forth?

Physics
2 answers:
agasfer [191]3 years ago
6 0
The correct answer is Longitudinal wave:)
vfiekz [6]3 years ago
3 0

Longitudinal wave

And this I know ;)

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BRAINLIST!!!
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  • <em><u>QB=-8×10^-9C.</u></em>
  • <em><u>QB=-8×10^-9C.We fixed balls A and B 5cm apart. Given e=1.6×10^-9C and k=9×10^9 S.I unit.</u></em>
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3 years ago
The process by which hot and cold air are transferred in the atmosphere is
elena-14-01-66 [18.8K]
Hey there!

<span>The process by which hot and cold air are transferred in the atmosphere is convection.</span>
8 0
3 years ago
The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb
balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = F_1=5.6 lb

Area of the piston = A_1=0.6 inches^2

Force applied on the brakes = F_2

Area of the brakes = A_2=1.8 inches^2

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

\frac{F_1}{A_2}=\frac{F_2}{A_2}

F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb

16.8 lb is the force on the brake pad of one wheel.

5 0
3 years ago
How much in kg is 1 N
Nookie1986 [14]
1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s2. 1 N (Earth) = 0.101971621297793 kg.
3 0
3 years ago
One of the main factors driving improvements in the cost and complexity of integrated circuits (ICs) is improvements in photolit
nika2105 [10]

Answer:

0.000003782 m

0.000001891 m

0.000001197125 m

Explanation:

\lambda = Wavelength = 248 nm

D = Diameter of beam = 1 cm

f = Focal length = 0.625 cm

The angle is given by

\theta=\dfrac{1.22\lambda}{D}

The width is given by

d=2\theta f\\\Rightarrow d=2\dfrac{1.22\lambda f}{D}\\\Rightarrow d=2\dfrac{1.22\times 248\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.000003782\ m

The required width is 0.000003782 m

Minimum resolvable line separation is given by

\dfrac{0.000003782}{2}=0.000001891\ m

The minimum resolvable line separation between adjacent lines is 0.000001891 m

when \lambda=157\ nm

d=2\dfrac{1.22\times 157\times 10^{-9}\times 6.25\times 10^{-2}}{1\times 10^{-2}}\\\Rightarrow d=0.00000239425\ m

The new minimum resolvable line separation between adjacent lines is

\dfrac{0.00000239425}{2}=0.000001197125\ m

6 0
3 years ago
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