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Hunter-Best [27]

2 years ago

7

Question: An increased number of rod cells increase the ability to see, A. during the day B. at night C. underwater or D. colors

?

2 answers:

Fynjy0 [20]2 years ago

8 0

B- at night

In eyes of the humans, and many other animals there are two types of receptors, the rods, and the cones. The cones have an important role in color vision and they possess different types of pigment which enable color vision.
On the other hand, the rods are more sensitive to light, and they are responsible for black and white vision.
Therefore, an increase in the number of rods can enable the animal to see betther in the dark.

bearhunter [10]2 years ago

4 0

The answer is B. at night

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A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

2 years ago

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

6 0

2 years ago

Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle o

Kazeer [188]

Answer:

2.2 µm

Explanation:

For constructive interference, the expression is:

$d\times sin\theta=m\times \lambda$

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle, $\theta$ = 15.8°

First bright fringe means , m = 1

So,

$d\times sin\ 15.8^0=1\times \597\ nm$

$d\times 0.2723=1\times \597\ nm$

$d=2192.43481\ nm$

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

<u>Distance between slits ≅ 2.2 µm</u>

8 0

3 years ago

Read 2 more answers

A 60 kg hockey player skating at 20 m/s gets slammed with a force of 950 N. How long does it take him to stop completely?

irakobra [83]

Answer:

2.4

Explanation:

8 0

2 years ago

liubo4ka [24]

Answer:

D

Explanation:

5 0

2 years ago