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3 years ago

What is the similarity and the principal difference between a beam of X-rays and a beam of light?

Physics

1 answer:

kati45 [8]3 years ago

3 0

Explanation:

Both are part of a family in physics which scientists refer to as the electromagnetic spectrum. All waves in this spectrum including light and x-rays travel at the same speed (3*10^8 m/s in a vacuum). However they differ in their wavelength and frequencies through the equation-----speed of light=wavelength multiplied by frequency.

Light may have a higher frequency than x-rays but lower wavelength. Either way they both multiply to receive the speed of light as the answer.

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Which statement is true for both types of transistors?

lara31 [8.8K]

For both NPN and PNP this is true:

The base is between the collector and the emitter.

8 0

3 years ago

Read 2 more answers

A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k

Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

7 0

2 years ago

What is the magnitude of the velocity when the elastic potential energy is equal to the kinetic energy? (Assume that U=0 at equi

zhannawk [14.2K]

Answer:

Explanation:

General Equation of SHM is given by

$x=A\cos \omega t$

$v=-A\omega \sin \omega t$

where x=position of particle

A=maximum Amplitude

$\omega =$ angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. $\frac{1}{2}kA^2$

where k=spring constant

Potential Energy is given by $U=\frac{1}{2}kx^2$

also it is given that Potential Energy(U) is equal to Kinetic Energy(K)

Total Energy $=K+U$

Total $=2U=2\times \frac{1}{2}kx^2$

$\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2$

$x=\pm \frac{A}{\sqrt{2}}$

at $x=\frac{A}{\sqrt{2}}$

velocity is $v=\frac{A\omega}{\sqrt{2}}$

6 0

3 years ago

Work done in taking charge from one point of a conductor to is another point is called

Yuliya22 [10]

Answer:

⁸

Explanation:

electric potential

I think so

6 0

2 years ago

A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?

adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

$\sum F = 0$

$F_b -W = 0$

$F_b = W$

$F_b = mg$

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

$\rho_w V_{displaced} g = mg$

Remember the expression for which you can determine the relationship between mass, volume and density, in which

$\rho = \frac{m}{V} \rightarrow m = V\rho$

In this case the density would be that of the object, replacing

$\rho_w V_{displaced} g = V\rho g$

Since the displaced volume of water is 0.429 we will have to

$\rho_w (0.429V) = V \rho$

$0.429\rho_w= \rho$

The density of water under normal conditions is $1000kg / m ^ 3$ , so

$0.429(1000) = \rho$

$\rho = 429kg/m^3$

The density of the object is $429kg / m ^ 3$

7 0

2 years ago