Donna is leaming about the water cycle in school. She knows that runoff is water that flows over land surfaces,

3. The electric field of a sinusoidal electromagnetic wave has an amplitude of 5.0 V/m. How much radiation energy passes through

True [87]

Answer:

e) $179$ J

Explanation:

$E_{o}$ = Magnitude of electric field = 5 V/m

$A$ = Area of window = 1.5 m²

$c$ = speed of electromagnetic wave = 3 x 10⁸ m/s

$\Delta t$ = time interval = 1 h = 3600 sec

radiation energy is given as

$U = (0.5)\epsilon _{o}E_{o}^{2}cA\Delta t$

$U = (0.5)(8.85\times 10^{-12})(5)^{2}(3\times 10^{8})(1.5)(3600)$

$U = 179$ J

6 0

2 years ago

Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such

Lyrx [107]

This question is incomplete, the complete question is;

- Calculate the difference in blood pressure between the feet and top of the head of a person who is 1.80m Tall

- Consider a cylindrical segment of a blood vessel 2.70 cm long and 3.10 mm in diameter. What additional outward force would such a vessel need to withstand in the person's feet compared to a similar vessel in her head

Answer:

- the difference in blood pressure is 18698.4 Pa

- the additional outward force F is 4.86 N

Explanation:

Given the data in the question;

we know that the expression for difference in blood pressure is;

ΔP = pgh

where p is density = 1060 kg/m³

g is acceleration due to gravity = 9.8 m/s²

and h is height = 1.80 m

now we substitute

ΔP = 1060 × 9.8 × 1.80

ΔP = 18698.4 Pa

therefore the difference in blood pressure is 18698.4 Pa

Also given that;

diameter of blood vessel d = 3.10 mm

radius r = 3.10 mm / 2 = 1.55 mm = 0.00155 m

length l = 2.70 cm = 0.027 m

Surface area of the cylindrical segment of a blood vessel is

A = 2πrl

we substitute

A = 2 × π × 0.00155 × 0.027

A = 2.6 × 10⁻⁴ m²

so

the required for will be;

F = PA

we substitute

F = 18698.4 Pa × 2.6 × 10⁻⁴ m²

F = 4.86 N

Therefore, the additional outward force F is 4.86 N

5 0

3 years ago

What is the wavelength of light (nm) that has a frequency of 6.44 x 1013 s-1?

lesantik [10]

4660

Hope this helped!
STSN

4 0

3 years ago

Read 2 more answers

When a 25-kg crate is pushed across a frictionless horizontal floor with a force of 200 N, directed 20 below the horizontal, th

Fofino [41]

Answer:

Option E is correct 310N

Explanation:

Given that the force used to push the crate is F = 200N

The force directed 20° below the horizontal

Mass of crate is m = 25kg

Weight of the crate can be determine using

W = mg

g is gravitational constant =9.8m/s²

W = 25×9.8

W = 245 N

Check attachment. For free body diagram and better understanding

Using newton second law along the vertical axis since we want to find the normal force

ΣFy = m•ay

ay = 0, since the body is not moving in the vertical or y direction

N—W—F•Sin20 = 0

N = W+F•Sin20

N = 245+ 200Sin20

N = 245 + 68.4

N = 313.4 N

The normal force is approximately 310 N to the nearest ten

3 0

3 years ago

A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl

LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

180 = θ2 + 90 + θ_p

θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection incident angle equals reflected angle,

ni sin θ_p = ns cos θ_p

n₂ / n₁ = sin θ_p / cos θ_p

n₂ / n₁ = tan θ_p

θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

θ_p = tan⁻¹ (1.33 / 1)

θ_p = 53.0º

n₂ = 1.40

θ_p = tan⁻¹ (1.40 / 1)

Tep = 54.5º

4 0

3 years ago