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Oksi-84 [34.3K]

3 years ago

5

__________ enable humans to see and detect the color of different objects.

2 answers:

Llana [10]3 years ago

6 0

The correct answer is C.

Aleks04 [339]3 years ago

4 0

The answer is C. Photoreceptors

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(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

2 years ago

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

2 years ago

Compared to the cells in a healthy body, Elisa’s cells are getting far fewer ____________________ molecules.

Elenna [48]

Answer:

glucose

Explanation:

7 0

2 years ago

Two rocks of different masses are rolling down a hill at the same speed. Which rock (small one or big one) would have more kinet

aivan3 [116]

Answer:

Kinetic energy of bigger rock will be more than that of smaller one.

Explanation:

Kinetic energy of the rock is given by,

Kinetic energy = $\frac{1}{2} m v^{2}$

As velocity of both the rocks are same. Thus, kinetic energy is directly proportional to the mass of the rock

Kinetic energy ∝ mass

So, For greater mass kinetic energy will be greater and for smaller mass kinetic energy will be smaller.

Hence, Kinetic energy of bigger rock will be more than that of smaller one.

3 0

2 years ago

Read 2 more answers

A planet orbits a star, in a year of length 3.37 x 107 s, in a nearly circular orbit of radius 1.04 x 1011 m. With respect to th

PIT_PIT [208]

Answer

Given,

Time period of star,T = 3.37 x 10⁷ s

Radius of circular orbit,R = 1.04 x 10¹¹ m

a) Angular speed of the planet

$\omega = \dfrac{2\pi}{T}=\dfrac{2\pi}{3.37\times 10^{7}}$

$\omega = 1.864\times 10^{-7}\ rad/s$

b) tangential speed

$v = r \omega = 1.04\times 10^{11}\times 1.864 \times 10^{-7}$

v = 1.94 x 10⁴ m/s

c) centripetal acceleration magnitude

$a = \dfrac{v^2}{r}= \dfrac{(1.94\times 10^4)^2}{1.04\times 10^{11}}$

a = 3.62 x 10⁻³ m/s²

8 0

2 years ago