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OLga [1]
3 years ago
5

If the wavelength (λ) of this wave is 0.012m, what is the speed (v) of this wave on graph A?

Physics
1 answer:
Natalka [10]3 years ago
7 0

Answer:

it would be 0.122 I think

Explanation:

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A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o
Tema [17]

Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

    W = P Δ V

    W = 1.013 x 10⁵ x (19.7 - 13.7)

    W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

  now,

 change in internal energy

  Δ U = Q - W

    Q = 3300 x 10³ cal

    Q = 3300 x 10³ x 4.186 J

    Q = 1.38 x 10⁷ J

now,

  Δ U = 1.38 x 10⁷  - 6.029 x 10⁵

  Δ U = 1.32 x 10⁷ J

6 0
3 years ago
The largest stars are _____ times the mass of the Sun. 10 100 1,000 10,000
Arte-miy333 [17]
<h3><u>Answer;</u></h3>

100 times

<h3><u>Explanation;</u></h3>
  • The largest stars are 100 times the mass of the Sun.
  • <u>The giant stars are about 10 to 100 times the radius of the sun</u>, which means they are 100 times brighter than the sun.
  • <em><u>The largest known star in terms of mass and brightness is known as the Pistol Star. It is believed to be 100 times as massive as our Sun, and 10,000,000 times as bright.</u></em>
5 0
3 years ago
What is the moving kinetic energy of a 1500kg car moving at 25m/s?
EleoNora [17]

Answer:

<h2>468,750 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 1500 \times  {25}^{2}  \\  = 750 \times 625

We have the final answer as

<h3>468,750 J</h3>

Hope this helps you

4 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
Anyone please help me with this question ASAP
Katyanochek1 [597]

Answer:

increases

Explanation:

it would have to work harder to get to two points together

7 0
3 years ago
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