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3 years ago

An x- ray is part of the electromagnetic spectrum. Compared to photons of visible light, an x-ray has?

Physics

2 answers:

Julli [10]3 years ago

8 0

Explanation:

X-rays are the electromagnetic waves whose photons are of higher energy and short wavelength that ranges beyond the visible region for the human-eyes.

The visible light lies between the wavelength of 4000 $A^{\circ}$ to 7000 $A^{\circ}$ .

The x-ray photons are the result of highly excited electron from its ground state than the electron releasing visible light photons.

The smaller atoms, in which the electron orbitals are separated by relatively low jumps in energy, are less likely to absorb X-ray photons.

bezimeni [28]3 years ago

3 0

Explanation:

X-rays are almost same as the visible light as they both electromagnetic waves with same speed. However, the only difference the energy level of photons of X- rays are higher than those of the visible light. This is because the frequency of the X-rays are higher.

E= $h\nu$

h= plank's constant and \nu is the frequency of the wave. So, higher the frequency higher will be the energy of the wave.

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Accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity

Crazy boy [7]

Answer:

The distance is

Explanation:

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Therefore, when

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⋅

and when

(

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⋅

Therefore,

The distance travelled in the fourth second is

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4 0

2 years ago

Two common ions of manganese are Mn2+ and Mn3+. Which binary ionic compound is likely to form? A) Mn2N3 B) MnBr4 C) MnCl D) MnO

meriva

Given that Oxygen has an oxidation state of 2 -, you can combine Mn 3+ with O 2- ions to form Mn2O3, and you can combine Mn 2+ with O 2- to form Mn2O2 which is MnO.

The other compounds imply oxidation states of N, Br and Cl that does not exist.

Therefore, the answer is the option D. MnO

4 0

3 years ago

Read 2 more answers

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$