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evablogger [386]
4 years ago
14

What is Quantum Physics?

Physics
1 answer:
vredina [299]4 years ago
8 0
Study of interaction between matter and energy using mechianical description
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How is gamma rays used to transfer energy?
expeople1 [14]
Gamma ray<span> (also called </span>gamma radiation<span>), denoted by the lower-case Greek letter </span>gamma,<span> is penetrating </span>electromagnetic radiation<span> of a kind arising from the </span>radioactive decay<span> of </span>atomic nuclei<span>. It consists of </span>photons<span> in the highest observed range of </span>photon energy<span>. </span>
6 0
4 years ago
A solid ball is released from rest and slides down a hillside that slopes downward at an angle 51.0 ∘ from the horizontal. what
Romashka [77]
In order for the object not to slip, the component of the weight parallel to the surface must be equal to the frictional force (which acts in the opposite direction):
F_{//}= F_a

The parallel component of the weight is:
F_{//} = mg \sin \alpha
where m is the object mass and \alpha is the angle of the inclined plane.

The frictional force is
F_a = \mu m g \cos \alpha
where \mu is the coefficient of static friction.

Equalizing the two forces, we have
mg \sin \alpha = \mu m g \cos \alpha
from which we find
\mu =  \tan \alpha

and so, in our problem the coefficient of static friction must be
\mu=\tan 51^{\circ} =1.23
4 0
3 years ago
.<br><br> 1. True False<br> In AM waves, amplitude changes.
Dennis_Churaev [7]
That is true Step by step:
3 0
3 years ago
Read 2 more answers
A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

6 0
3 years ago
Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,
Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0
4 years ago
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