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Feliz [49]
3 years ago
9

When two adjacent lights blink on and off in quick succession, we perceive a single light moving back and forth between them. th

is is called?
Physics
1 answer:
makkiz [27]3 years ago
5 0
This is called the Phi Phenomenon.
This is an illusion of movement created when two or more adjacent lights blink on and off in quick succession; when two adjacent stationary lights blink on and off in quick succession; we perceive a single light moving back and forth between them. It is an optical illusion of perceiving a series of still images, when viewed in rapid succession, as continuous motion. 
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What do you know about potential energy? And how do we use it?
Liula [17]
Energy is the ability to do work or cause change. There are basically two main types of energy, kinetic and potential. Potential energy is energy that is stored. There are various types of stored, or potential energy. Chemical energy from a battery is a potential form of energy, elastic energy in a stretched rubber band is a form of potential energy, but the most commonly referred to form of potential energy in physics is that of gravitational potential energy. This is energy that is stored due to an object's position. It is dependent on the mass of the object, the height of the object above the ground or Earth, and the acceleration due to gravity.
7 0
3 years ago
A roulette wheel with a 1.0m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions ( 220 rad ) after
Alex17521 [72]
Max ang. speed(u) = 18 rad/s
final ang. speed(v) = 0
ang. displacement(s) = 220 rad

ang. acceleration = (v^2 - u^2)/2s = -18^2 / 2*220 = -0.7364 rad/s^2

v = u +at
0 = 18 - 0.7364t
t = 18/0.7364
t = 24.44 seconds
4 0
3 years ago
Find the ratio of the final speed of the electron to the final speed of the hydrogen ion, assuming non-relativistic speeds. Take
KiRa [710]

Answer:

\frac{V_{e}}{V_{h}}=0.428*10^{2}

Explanation:

From conservation of energy states that

K_{i}+v_{i}=v_{f}+K_{f}\\ as\\K_{i}=0\\K_{f}=1/2mv^{2}\\ v_{i}=qv\\v_{f}=0\\So\\qv=1/2mv^{2}\\ v=\sqrt{\frac{2qv}{m} }\\ Velocity_{electron}=\sqrt{\frac{2qv}{m_{e}} }\\Velocity_{hydrogen}=\sqrt{\frac{2qv}{m_{h}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{\frac{2qv}{m_{e}}}{\frac{2qv}{m_{h}}}}\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{m_{h}}{m_{e}} }\\\frac{V_{e}}{V_{h}}=\sqrt{\frac{1.67*10^{-27} }{9.11*10^{-31} } }\\\frac{V_{e}}{V_{h}}=0.428*10^{2}

5 0
3 years ago
A reactor operating at 1 MW is scrammed by instantaneous insertion of $5.00 of negative reactivity. Approximately how long does
Novosadov [1.4K]

Answer:

time is 3333.33 min or 55.55 hr

Explanation:

given data

reactor operating = 1 MW

negative reactivity =  $5

power = 1 miliwatt

to find out

how long does it take

solution

we know here power coefficient that is

power coefficient = \frac{10^{6} }{10^{6} }

power coefficient = 1

so time required to reach power is

power =  reactivity × time / power coefficient + reactor operating

1 × 10^{-3} = -5 t / 1  +  1 × 10^{6}

5t =  10^{6} - 10^{-3}

t = 199999.99 sec

so time is 3333.33 min or 55.55 hr

4 0
3 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
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