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3 years ago

The persistence of vision for normal eye is

Physics

2 answers:

attashe74 [19]3 years ago

6 0

Answer:

The answer is 1/16

Explanation:

1. Persistence of vision refers to the optical illusion that occurs when visual perception of an object does not cease for some time after the rays of light proceeding from it have ceased to enter the eye. 2. The persistence of vision for normal eye is 1/16 if a second.

pashok25 [27]3 years ago

4 0

Answer:

idk

Explanation:

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Which of the following is a true statement about work?

Over [174]

Answer:

It measure the effort it takes for an object to move

Explanation:

Work done is the force applied to move a body through a particular distance.

It can also to be said to be a measure of the effort it takes for an object to move.

The unit of work done is in Joules.

To find work done, we use the expression below:

Work done = Force x distance

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2 years ago

Which of the following is not a valid conversion factor?

kykrilka [37]

The third option is wrong

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2 years ago

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2 years ago

One cycle of the power dissipated by a resistor ( R = 800 Ω R=800 Ω) is given by P ( t ) = 60 W , 0 ≤ t < 5.0 s P(t)=60 W, 0≤

OLga [1]

Answer:

42.5W

Explanation:

To solve this problem we must go back to the calculations of a weighted average based on the time elapsed thus,

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

We need to calculate the average power dissipated by the 800\Omega resistor.

Our values are given by:

$P(t)=60 W, 0\leq t$

$P(t)=25 W, 5.0\leq t$

Aplying the values to the equation we have:

$Power_{avg} = \frac{P_1(t_1)+P_2(t_2)}{t_1+t_2}$

$Power_{avg} = \frac{60(5-0)+25(10-5)}{(5-0)+(10-5)}$

$Power_{avg} = 42.5W$

5 0

2 years ago

An iron wire has length 8.0m and a diameter 0.50mm. The sir has a resistance R.

Rudik [331]

The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.

R = </span>ρ $\frac{l}{A}$
where l = length of the wire
A = area of the wire
A = $\pi r^{2}$ where, r = $\frac{diameter of wire}{2}$

Thus, on finding the ratio of resistance of the two wires, we get,

$\frac{R1}{R2} = \frac{l1A2}{l2A1}$

here, R1 = R
l1 = 8m
l2 = 2m
A1=π $0.25^{2}$
A1=π $0.50^{2}$

we get. R2 = 16R

7 0

3 years ago