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arlik [135]
3 years ago
10

The absorption of longwave radiation in the atmosphere is popularly called ________.

Physics
1 answer:
Sliva [168]3 years ago
4 0
I believe it is called the Greenhouse effect.
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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

5 0
3 years ago
The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t
Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

B = \frac{\mu_oI}{2\pi r}\\\\r =  \frac{\mu_oI}{2\pi B}

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\

<u>r = 0.05 m = 5 cm</u>

8 0
2 years ago
A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0
2 years ago
Which process do hydrogen atoms use to make the Sun's energy?
Maru [420]
Nuclear fusion, converts hydrogen atoms into helium
8 0
3 years ago
Graphite crystals can be used as a lubricant.<br> TRUE<br> FALSE
timurjin [86]
I think its true because they are a soft solid that can be broken down into a liquid... Hope this helps in any way :3
6 0
2 years ago
Read 2 more answers
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