All categories

2 years ago

If the Earth absorbs more insolation from the sun than it radiates back, the Earth _____________.

Physics

1 answer:

professor190 [17]2 years ago

8 0

Answer:

If the Earth absorbs more insolation from the sun than it radiates back, the Earth. warms

You might be interested in

How much conventional current must you run in a solenoid with radius = 0.05 m and length = 0.39 m to produce a magnetic field in

Ugo [173]

Answer:

Explanation:

radius of the solenoid, r = 0.05 m

length of the solenoid, l = 0.39 m

Magnetic field of the solenoid, B = 2 x 10^-5 T

Number of turns, N = 200

The magnetic field of the solenoid is given by

$B=\mu _{0}ni$

where, i be the current and n be the number of turns per unit length

n = N / l = 200 / 0.39 = 512.8

$2\times 10^{-5}=4 \times 3.14\times 10^{-7}\times 512.8\times i$

i = 0.031 A

3 0

2 years ago

14. Which one of the following pictures shows the object that is the most dense? *

rewona [7]

Answer:

Explanation:

Density is about how closely compact molecules are. (^-^)

5 0

2 years ago

Mr. Smith used 606 kWh for the month. If the service charge was $61.37 (see back page of the bill), what is the approximate char

Murrr4er [49]

Answer:

.10/KWh

Explanation:

divide 606 by 61.37 and you get .1012...

7 0

2 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

4 0

3 years ago

In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg

harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 × $10^{-2}$ m

separation L = 0.5 mm = 0.5 × $10^{-3}$ m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 × $10^{-2}$ / 11

θ = 0.234

and we know diffraction grating for max

L sinθ = mλ

here we know m = 1 so put all value and find λ

L sinθ = mλ

0.5 × $10^{-3}$ sin(0.234) = 1 λ

λ = 2042.02 × $10^{-9}$ m

λ = 2042 nm

3 0

3 years ago