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1 year ago

8

When developing a model of the cycling of water between the land, the ocean, and the atmosphere, one must include the forces tha

t propel the water. For the model, these forces would be-

1 answer:

zhannawk [14.2K]1 year ago

7 0

Answer: gl

Explanation:

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A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s

Furkat [3]

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

$p_{rad} = \frac{I}{c}$

<u>Where:</u>

$p_{rad}$ : is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

$p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}$

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

3 0

2 years ago

Read 2 more answers

In ocean waves, water particles move ________ and energy moves ________.

morpeh [17]

In ocean waves, water particles move with mechanical energy and energy moves with gravity

Not sure but hope it helps!

6 0

1 year ago

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a film of transparent material 120 nm thick and having refractive index 1.25 is placed on a glass sheet having refractive index

Eva8 [605]

Answer:

a) 600nm

b) 300nm

Explanation:

the path difference = 2t

t = thickness of the film

L' = wavelength of light in film = L/n

L = wavength of light in air

n = refractive index of glass

(a)

for destructive interference 2t = L'/2 = L/2n

L = 4*t*n

= 4*120*10^-9*1.25

L = 600 nm

(b)

for constructive interference 2t = L' = L/1.25

L = 2tn

= 2 × 1.25 × 120nm

= 300 nm

4 0

2 years ago

Honeybees can see light in the ________ range of the electromagnetic spectrum.

konstantin123 [22]

Humans can see wavelengths in the visible part of the electromagnetic spectrum. That is the range of approximately 400 - 700 nm. Honeybees can see visible light and about 100 nm more in the ultraviolet part of the electromagnetic spectrum. That is approximately 300 - 700 nm.

4 0

2 years ago

1. California sea lions can swim as fast as 40.0 km/h. Suppose a sea lion begins to chase a fish at this speed when the fish is

Pie

#1

In order to chase the fish the distance traveled by sea lion in time t must be equal to the distance of sea lion from the fish and distance traveled by fish in the same time.

So here we can say let say sea lion chase the fish in time "t"

then here we have

$d_1 = d_2 + L$

here

d1 = distance covered by sea lion in time t

d2 = distance covered by fish in the same time t

L = distance between fish and sea lion initially = 60 m

$d_1 = v_1 * t$

$d_1 = (40*\frac{5}{18})*t = \frac{100}{9}*t$

$d_2 = (16*\frac{5}{18})*t = \frac{40}{9}*t$

$\frac{100}{9}*t = \frac{40}{9}*t + 60$

$\frac{100}{9}*t - \frac{40}{9}*t = 60$

$\frac{60}{9}*t = 60$

$t = 9 s$

So it will take 9 s to chase the fish by sea lion

# 2

velocity of truck on road = 25 m/s along North

velocity of dog inside the truck = 1.75 m/s at 35 degree East of North

$v_{dt} = 1.75 cos35\hat j + 1.75sin35 \hat i$

$v_{dt} = 1.43 \hat j + 1 \hat i$

we can write the relative velocity as

$v_d - v_t = 1.43 \hat j + 1 \hat i$

$v_d = v_t + (1.43 \hat j + 1 \hat i)$

now plug in the velocity of truck in this

$v_d = 25 \hat j + (1.43 \hat j + 1 \hat i)$

$v_d = 26.43 \hat j + 1 \hat i$

so it is given as

$v_d = \sqrt{26.43^2 + 1^2} = 26.44 m/s$

direction will be given as

$\theta = tan^{-1}\frac{v_x}{v_y}$

$\theta = tan^{-1}\frac{1}{26.43} = 2.2 degree$

so with respect to ground dog velocity is 26.44 m/s towards 2.2 degree East of North

8 0

2 years ago

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