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Which scientific phenomenon helps scientists determine the chemical composition of some stars?

ser-zykov [4K]

The correct option is DISPERSION OF LIGHT AS IT ENTERS EARTH'S ATMOSPHERE. Scientists use astronomical spectrometer to analyse the composition of the sun and the stars. Astronomical spectrometer works on the basis of dispersion of light as it enters the earth's atmosphere.

6 0

3 years ago

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A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were f

STatiana [176]

Answer:

y = 1.19 m and λ = 8.6036 10⁻⁷ m

Explanation:

This is a slit interference problem, the expression for destructive interference is

d sin θ = m λ

indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m

λ = d sin θ / m

let's calculate

λ = 4.50 10⁻⁶ sin 35 /3

λ = 8.6036 10⁻⁷ m

for the separation distance from the central stripe, we use trigonometry

tan θ= y / L

y = L tan θ

the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits

L = 2 -0.30

L = 1.70 m

let's calculate

y = 1.70 tan 35

y = 1.19 m

3 0

3 years ago

What are the parts of a wedge and their function?

Svet_ta [14]

Answer:

A wedge is a machine that consists of two inclined planes, giving it a thin end and thick end. A wedge is used to cut or split apart objects. Force is applied to the thick end of the wedge, and the wedge applies force to the object along both of its sloping sides. This force causes the object to split apart

5 0

3 years ago

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

Now the resultant of the two velocities perpendicular to each other:

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

2)

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

3)

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force

aliina [53]

Answer:

v_f = 10.38 m / s

Explanation:

For this exercise we can use the relationship between work and kinetic energy

W = ΔK

note that the two quantities are scalars

Work is defined by the relation

W = F. Δx

the bold are vectors. The displacement is

Δx = r_f -r₀

Δx = (11.6 i - 2j) - (4.4 i + 5j)

Δx = (7.2 i - 7 j) m

W = (4 i - 9j). (7.2 i - 7 j)

remember that the dot product

i.i = j.j = 1

i.j = 0

W = 4 7.2 + 9 7

W = 91.8 J

the initial kinetic energy is

Ko = ½ m vo²

Ko = ½ 2.0 4.0²

Ko = 16 J

we substitute in the initial equation

W = K_f - K₀

K_f = W + K₀

½ m v_f² = W + K₀

v_f² = 2 / m (W + K₀)

v_f² = 2/2 (91.8 + 16)

v_f = √107.8

v_f = 10.38 m / s

3 0

3 years ago