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2 years ago

What is the intensity of sunlight on earth.

Physics

1 answer:

azamat2 years ago

4 0

About 1,360 watts per square meter

At Earth's average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions.

You might be interested in

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0

2 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

3 years ago

A cylindrical storage tank has a radius of 1.35 m. When filled to a height of 3.45 m, it holds 14,014 kg of a liquid industrial

Kruka [31]

Answer:

709.93 kg/m³

Explanation:

Density: This can be defined as the ratio of the mass of a body to the volume of that body. The S.I unit of density is kg/m³

From the question above,

D = m/v.............................. Equation 1

Where D = density, m = mass, v = volume.

Note: The volume of the liquid is equal to the volume of the height occupied by the liquid in the container

Since the tank is cylindrical,

v = πr²h........................ Equation 2

Where r = radius of the the tank, h = height of the liquid in the tank

Substitute equation 2 into equation 1

D = m/(πr²h)............... Equation 3

Given: m = 14014 kg, r = 1.35 m, h = 3.45 m, π = 3.14

Substitute into equation 3

D = 14014/(3.14×1.35²×3.45)

D = 14014/19.74

D = 709.93 kg/m³

6 0

3 years ago

Which of the following best describes the upper respiratory tract?

In-s [12.5K]

A). It takes air in from outside the body.

3 0

3 years ago

Please solve the Problem.

STatiana [176]

(a) For series circuit, current in 14 ohms = current in 72 ohms = 0.698 A

(b) Power loss in each series resistor, 14 ohms = 6.82 W and 72 ohms = 35.1 W.

(d) Power loss in each parallel resistor, 14 ohms = 257.4 W and 72 ohms = 49.8 W.

<h3>Current in each series resistors</h3>

The total resistance = R1 + R2

R = 14 + 72 = 86 ohms

Current = V/R = 60/86 = 0.698 A

Since the resistors are in series, current in 14 ohms = current in 72 ohms = 0.698 A

<h3>Power loss in each series resistor</h3>

P = I²R

P(14 ohms) = (0.698)² x 14 = 6.82 W

P(72 ohms) = (0.698)² x 72 = 35.1 W

<h3>Current in each resistor for parallel arrangement</h3>

Total resistance, 1/R = 1/R1 + 1/R2

1/R = 1/14 + 1/72

1/R = 0.0853

R = 1/0.0853

R = 11.72

Total current in the circuit = V/R = 60/11.72 = 5.12 A

Current in 14 ohms = 60/14 = 4.29 A

Current in 72 ohms = 60/72 = 0.83 A

<h3>Power loss in each parallel resistor</h3>

P = IV

P(14 ohms) = (4.29) x 60 = 257.4 W

P(72 ohms) = 0.83 x 60 = 49.8 W

Learn more about current here: brainly.com/question/24858512

#SPJ1

5 0

2 years ago