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3 years ago

8

When is it necessary to use relativistic quantum mechanics?

1 answer:

Ugo [173]3 years ago

4 0

Answer:

To see objects smaller than microscopic limits

Explanation:

The theory of Relativistic Quantum mechanics can be applied to particles that are massive and propagates at all velocities even those which are comparable to the speed of light and is capable to accommodate particles that are mass less. This theory find its application in atomic physics, high energy physics, etc.

It is necessary to use relativistic quantum mechanics when it is desired to see the objects that are too small to be seen with the help of microscope.

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Why would the planets move in a straight path if there was no gravitational energy from the sun

omeli [17]

Explanation:

The sun's gravitational force is very strong. If it were not, a planet would move in a straight line out into space. The sun's gravity pulls the planet toward the sun, which changes the straight line of direction into a curve. This keeps the planet moving in an orbit around the sun

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3 years ago

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g In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constr

Lorico [155]

To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.

In other words the acceleration can be described as

$a = \frac{GM}{r^2}$

Where

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius of Earth

This equation can be differentiated with respect to the radius of change, that is

$\frac{da}{dr} = -2\frac{GM}{r^3}$

$da = -2\frac{GM}{r^3}dr$

At the same time since Newton's second law we know that:

$F_w = ma$

Where,

m = mass

a =Acceleration

From the previous value given for acceleration we have to

$F_W = m (\frac{GM}{r^2} ) = 600N$

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

$dF_W = -2(m\frac{GM}{r^2})(\frac{dr}{r})$

$dF_W = -2F_W(\frac{dr}{r})$

But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

$dF_W = -0.3N$

Therefore there is a weight loss of 0.3N every kilometer.

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4 years ago

A light year is _____.

777dan777 [17]

Alight year is the distance that ligth travels in a year.

Answer: C. The distance tat light travels in a year.

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3 years ago

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The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a sp

AlekseyPX

Answer:

$t = 0.93 s$

Part b)

$d = 3.98 m$

Part c)

$v_x = 4.28 m/s$

$v_y = -11.49 m/s$

Explanation:

The two components of the velocity of the ball is given as

$v_x = 4.9 cos29$

$v_x = 4.28 m/s$

$v_y = 4.9 sin29$

$v_y = 2.37 m/s$

Part a)

now we know that the displacement in y direction is given as

$\Delta y = 6.4 m$

so we have

$\Delta y = v_y t + \frac{1}{2}gt^2$

$6.4 = 2.37 t + 4.9 t^2$

$t = 0.93 s$

Part b)

Distance of the ball in x direction of the motion is given as

$d = v_x t$

$d = 4.28 \times 0.93$

$d = 3.98 m$

Part c)

In x direction the velocity will remain the same always

$v_x = 4.28 m/s$

while in Y direction we can use kinematics

$v_y = v_{oy} + at$

$v_y = -2.37 - 9.81(0.93)$

$v_y = -11.49 m/s$

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3 years ago

The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?

ICE Princess25 [194]

Answer:

660V

Explanation:

V=IR

V=?,I=11A,R=60w

V=60×11

=660V

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3 years ago