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IceJOKER [234]
4 years ago
14

How is light being described?

Physics
1 answer:
Slav-nsk [51]4 years ago
7 0
Light is described as both wave as well particle because light behave as wave in phenomenon like diffraction, interference, refraction, Doppler effect, etc whereas it behave as a particle in phenomenon like photoelectric effect, etc. Hope this helps.
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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
Protect your plants against ________ off disease​
Kazeer [188]

Answer:

attacking

!!!!!! !!!!!

5 0
3 years ago
Read 2 more answers
Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0
4 years ago
Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given tha
pshichka [43]
For resistance we have R=ρ l/a
 thus for conductance we have K=σ a/l
conductance,K=1/R
conductivity,σ =1/ρ

σ = .80 Ω-1 cm-1
l =9 cm
a = 3 cm²
K=.80 ×3/9
  =0.26 Ω-1


6 0
3 years ago
Read 2 more answers
What should be the indication on the magnetic compass as you roll into a standard rate turn to the right from a northerly headin
lapo4ka [179]

Answer:

The compass will indicate a turn to the left.

Explanation:

The magnetic compass has a needle pointing towards north. It is used for navigation and it shows direction with respect to the geographical cardinal direction.

As we move to the right of the northerly heading in northern Hemisphere, the direction of needle will move towards left but with a faster rate as now the direction of north is no his left.

This shows the compass if following its direction properly.

4 0
3 years ago
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