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3 years ago

Which is example of radiation

Physics

2 answers:

emmainna [20.7K]3 years ago

6 0

Answer:

Ultraviolet light from the sun.

Explanation:

This is an example of radiation.

SSSSS [86.1K]3 years ago

6 0

Answer:

X-Ray

Explanation:

x-Ray is an example of radiation.

You might be interested in

What is the calculation for elastic potential?

grin007 [14]

Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.

3 0

3 years ago

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibra

Norma-Jean [14]

Answer:

1456 N

Explanation:

Given that

Frequency of the piano, f = 27.5 Hz

Entire length of the string, l = 2 m

Mass of the piano, m = 400 g

Length of the vibrating section of the string, L = 1.9 m

Tension needed, T = ?

The formula for the tension is represented as

T = 4mL²f²/ l, where

T = tension

m = mass

L = length of vibrating part

F = frequency

l = length of the whole part

If we substitute and apply the values we have Fri. The question, we would have

T = (4 * 0.4 * 1.9² * 27.5²) / 2

T = 4368.1 / 2

T = 1456 N

Thus, we could conclude that the tension needed to tune the string properly is 1456 N

4 0

3 years ago

A viscous fluid is flowing through two horizontal pipes. They have the same length, although the radius of one pipe is three tim

Rudik [331]

The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,

$Q_1 = Q_2$

The definition of flow is given by

$Q =VA$

Where

V = Velocity

A = Area

The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.

In this way

$Q_A = Q_B$

$Q_B = 2.94*10^2 m^3/s$

7 0

3 years ago

A light beam travels at 1.94×108 in quartz. The wavelength of the light in quartz is 355 .Part AWhat is the index of refraction

Alja [10]

A) 1.55

The speed of light in a medium is given by:

$v=\frac{c}{n}$

where

$c=3\cdot 10^8 m/s$ is the speed of light in a vacuum

n is the refractive index of the material

In this problem, the speed of light in quartz is

$v=1.94\cdot 10^8 m/s$

So we can re-arrange the previous formula to find n, the index of refraction of quartz:

$n=\frac{c}{v}=\frac{3\cdot 10^8 m/s}{1.94\cdot 10^8 m/s}=1.55$

B) 550.3 nm

The relationship between the wavelength of the light in air and in quartz is

$\lambda=\frac{\lambda_0}{n}$

where

$\lambda$ is the wavelenght in quartz

$\lambda_0$ is the wavelength in air

n is the refractive index

For the light in this problem, we have

$\lambda=355 nm\\n=1.55$

Therefore, we can re-arrange the equation to find $\lambda_0$ , the wavelength in air:

$\lambda_0 = n\lambda=(1.55)(355 nm)=550.3 nm$

4 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago