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Dimas [21]
3 years ago
10

"Close the door! We aren't heating the outdoors!" is a phrase my father used, especially in the winter. As a matter of fact, all

energy consumption eventually turns into heat, which becomes part of Earth's heat. Wikipedia gives the total energy consumption in 2008 to be 15 TW (15 terawatts = 15 x 1012 W). We can spread this power out over the entire surface area of the Earth to calculate the radiative forcing equivalent of the direct use of energy to be _____ W/m2. The radiating area of the Earth is 4Ï R2, where the radius of the Earth is about 6366 km. (watch your unit conversions).
Physics
1 answer:
Alina [70]3 years ago
7 0

Answer:

I = 2,945 10⁻² W / m²

Explanation:

In this exercise it is requested to calculate the intensity of radiation that is defined by the power per unit of ares

           I = P / A

Approach the Earth as a sphere, with area

         A = 4π R²

Let's replace

           I = P / 4π R²

Let's reduce the magnitudes

      W = 15 10¹² W

       R = 6366 km (10³ m / 1km) = 6.366 10⁶ m

Let's calculate

       I = 15 10¹²/4π (6.366 10⁶)²2

       I = 2,945 10⁻² W / m²

This is a very small amount compared to the intensity coming from the Sun 1353 W / m²

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A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part
marta [7]

Answer:

Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2y i + 1.2x j)

So comparing comparing coefficients

let compare x axis component

-( 3.30×10−7N )i=-1.2qy i

−3.30×10−7N = -1.2qy

y= -3.3×10^-7/-1.2q

y= -3.3×10^-7/-1.2×-5.10×10^-9)

y=-53.92m/s

Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cos

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

6 0
2 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the lo
Zepler [3.9K]

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

\phi = BAcos \theta

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb

Hence the magnetic flux Φ through the loop is 0.5849Weber

5 0
2 years ago
The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re
gogolik [260]
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

7 0
3 years ago
A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s
earnstyle [38]

Answer:

0.4rad/s²

Explanation:

Angular acceleration is the time rate of change of angular velocity . In SI units, it is measured in radians per second squared (rad/s²)

w1 = 4rad/s, w2 =2rad/s, t = 5sec, r = 0.30m

a = ∆w/t

a = (w2 - w1)/t

a = (2 - 4)/5 = -2/5 =

a = - 0.4rad/s²

The -ve sign indicates a deceleration in the motion

Good luck

3 0
2 years ago
If a force of 450N pushes a box 4 meters in 2 seconds, how much power was developed?
strojnjashka [21]

Work = (force) x (distance) = (450 N) x (4 m) = 1,800 joules

Power = (work) / (time) = (1,800 joules) / (2 sec) = 900 watts .
5 0
3 years ago
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