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3 years ago

compare to visible light, the wavelength of x-rays is shorter longer or the same and the frequency is lower higher or the same

Physics

1 answer:

Doss [256]3 years ago

7 0

Answer:shorter higher

Explanation:Compare to visible light, the wavelength of X-rays is shorter and then frequency is higher

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A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top

Sveta_85 [38]

Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.

Given,

Length, L = 4 m

Outer diameter, D = 300mm, D= 0.3 m

Thickness, t = 50 mm, t = 0.05 m

Stress produced, σ = 75000 kN/m²

Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²

Calculating the diameter of the cylinder,

Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)

d= 0.2 m

(i) Magnitude of the load P:

Using the relation, σ =P/A

P = σ × A = 75000 × π /4 (D² – d² )

P= 75000 × π/4 (0.3² – 0.2²)

P= 75000 × π/4 (0.09 – 0.04)

P = 2945.2 kN

Hence, Magnitude of the load P is 2945.2 kN.

(ii) Longitudinal strain produced, e :

Using the relation, Strain, (e) = stress/E

e= 75000/(1.5 x 10⁸)= 0.0005

Hence, the Longitudinal strain produced is 0.0005.

(iii)Total decrease in length, dL:

The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.

Strain = change in length/original length

e= dL/L

0.0005 = dL/4

dL = 0.0005 × 4m = 0.002m=2mm

Hence,the decrease in length is 2 mm.

Learn more about Elasticity here:

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5 0

2 years ago

What is a example of a real life scientific inquiry problem

Papessa [141]

What is an example of how you can use scientific inquiry to solve a real life problem.

8 0

3 years ago

1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

109.6 cm , - 1.74 , real

1.5

Explanation:

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

3 years ago

A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee

Olegator [25]

Quite low

Gravitation force = $\frac{6.67* 10^{-11}*0.145*6.8 }{ 0.5^{2} }$ = 26.3 * $10^{-11}$ = 2.63 * $10^{-10}$ N

6 0

3 years ago

Read 2 more answers

1. Ramon puts two magnetic toy trains very close to each other on a track. What will happen next, and

FrozenT [24]

Answer:

If one side of the train is positive and the other is negative they will attract if they are the same then they will repel.

Explanation:

If both are positive they will repel if both are negative they will repel and if they are opposites they will attract.

5 0

3 years ago

Read 2 more answers