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4 years ago

If one is missing the opsin which detects wavelengths of approximately 560 nm what color would they be unable to see?

Physics

1 answer:

Licemer1 [7]4 years ago

4 0

As per the question the wavelength of color of light is given as 560 nm.

We are asked to determine the color corresponding to this wavelength.

The visible spectrum corresponds to the wavelength range of 380 nm to 750 nm.

The visible spectrum contains seven colors of light.

The green light corresponds to the wavelength range of 495 nm to 570 nm.

Hence the correct answer to this question will be green light.

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two parallel plates of area 5.68*10^-4 m^2 have equal and opposite charges of 8.38*10^-11 C placed on them. what is the electric

shutvik [7]

Answer: unit n/c

Explanation: This should be the answer because I just asked my Physics Teachers I hope this is correct

Hope This Helps :)

4 0

3 years ago

Read 2 more answers

A bowling ball of mass 4 kg moves in a straight line at 3.93 m/s. How fast must a Ping-Pong ball of mass 2.293 g move in a strai

Nonamiya [84]

M = mass of the bowling ball = 4 kg

V = speed of bowling ball = 3.93 m/s

P = magnitude of momentum of bowling ball = ?

magnitude of momentum of bowling ball is given as

P = MV

inserting the values

P = 4 x 3.93

P = 15.72 kgm/s

m = mass of ping-pong ball = 2.293 g = 2.293 x 10⁻³ kg

v = speed of the ping-pong ball = ?

p = magnitude of momentum of ping-pong ball

Given that :

magnitude of momentum of ping-pong ball = magnitude of momentum of bowling ball

p = P

m v = 15.72

(2.293 x 10⁻³) v = 15.72

v = 6.86 x 10³ m/s

4 0

3 years ago

What is the resultant of 5N force pointing north and 7N force pointing south? Do not forget include direction

balandron [24]

Given forceF

=5N and F

=7N and θ=60

We know resultant force F=

+2(5)(7)cos60

25+49+35

109

it's your answer

4 0

3 years ago

When a Lunar Module landed on the Moon, it used thrusters to slow its descent to the surface. When other spacecraft are returned

katen-ka-za [31]

They had to use thrust because there is no air on the moon meaning that parachutes would be totally useless because parachutes require air to create drag. Fun fact since there is no air on the lunar surface sound waves cannot be made since there are no air molecules for the waves to pass through, so that's why they use radio's in space.

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3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$