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3 years ago

When the mirror is rotated, the normal will turn as well, but will the incident Ray and reflected ray turn?

1 answer:

6 0

When a mirror is rotated . . .

-- The incident ray doesn't turn. It's just the line from the source to the mirror.
It would be there, in the same place, even if there was no mirror.

-- The normal turns. It's the line perpendicular to the mirror, so it must turn
with the mirror.

-- Since the normal tuns and the incident ray doesn't, the angle between them
must change. And since the angle of the reflected ray is equal to the angle of
the incident ray, the reflected ray must also turn.

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Where would a new neuron come from

erica [24]

Neurogenesis does not occur everywhere in the brain but is evident in the hippocampus and olfactory bulb and perhaps in the cerebral cortex. New neurons are born not from mature nerve cells but rather develop from neural stem cells that remain in our brains throughout life.

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3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

4 0

3 years ago

A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current

larisa [96]

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega$

And now we can use Ohm's law to find the current in the circuit:

$V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A$

6 0

3 years ago

Three masses are located in the x- y plane as follows: a mass of 6 kg is at (0 m, 0 m), a mass of 4 kg is at (3 m, 0 m), and a m

enot [183]

Answer:

The center of mass of three mass in the x-y plane is located at (1,0.5).

Explanation:

It is given that, a mass of 6 kg is at (0,0), a mass of 4 kg is at (3,0), and a mass of 2 kg is at (0,3). We need to find the center of mass of the system. Center of mass in x direction is :

$C_x=\dfrac{6\times 0+4\times 3+2\times 0}{6+4+2}\\\\C_x=1$

The center of mass in y direction is :

$C_y=\dfrac{6\times 0+4\times 0+2\times 3}{6+4+2}\\\\C_y=0.5$

So, the center of mass of three mass in the x-y plane is located at (1,0.5).

3 0

3 years ago

An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result

mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

$s=ut+\frac{1}{2}gt^{2}$

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

$s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}$

So the distance covered by the acorn is 3.136 m.

8 0

3 years ago