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sweet-ann [11.9K]
2 years ago
5

When Earth’s Northern Hemisphere is tilted toward the Sun during June, some would argue that the cause of our seasons is that th

e Northern Hemisphere is physically closer to the Sun than the Southern Hemisphere, and this is the primary reason the Northern Hemisphere is warmer. What argument or line of evidence could contradict this idea?
Physics
1 answer:
Ludmilka [50]2 years ago
4 0

Answer:

Distance of Earth from the Sun has nothing to do with the seasons only the tilt is responsible for the change in seasons.

Explanation:

The Earth's tilt does cause the seasons but the distance from the sun and has nothing to do with the change in seasons. In June, when the Northern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere summer the Earth is actually farthest from the Sun. In January, when the Southern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere winter the Earth is actually closest to the Sun. This is caused due to the elliptical orbit of the Earth. So, distance of Earth from the Sun has nothing to do with the seasons.

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In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
Rus_ich [418]
M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
   = (5.60 − 2)/(2 + 5.60) x 9.81 
   = = 4.65m/s^2
5 0
2 years ago
A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When t
klasskru [66]

Answer:

a.a. \ \frac{7}{25}

b.\ \ \ P(D_1D_2)=\frac{6}{25}

Explanation:

a. Find the probability that exactly one photo detector of a pair is acceptable:

Let A_i=i^{th} photo is accepted and the probability D_i=i^{th} is defected.

Therefore:

P(A_i)=3/5,\ P(A_2|A_1)=4/5,\ \ P(A_2|D_1)=2/5\\\\\\=P(A_1D_2)+P(D_1A_2)\\\\=\frac{3}{5}\times\frac{1}{5}+\frac{2}{5}\times\frac{2}{5}\\\\=\frac{7}{25}

#The probability of exactly one photo detector of a pair is accepted is 7/25

b.Find the probability that both photo detectors in a pair are defective,P(D1D2):

P(D_1D_2)=\frac{2}{5}\times \frac{3}{5}\\\\=\frac{6}{25}

Hence, from out tree diagram,the probability that both photo detectors in a pair are defective is 6/25

4 0
2 years ago
You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just
Zepler [3.9K]

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

4 0
2 years ago
When Jim and Rob ride bicycles, Jim can only accelerate at three-quarters the acceleration of Rob. Both startfrom rest at the bo
Natali5045456 [20]

Answer:

46.4 s

Explanation:

5 minutes = 60 * 5 = 300 seconds

Let g = 9.8 m/s2. And \theta be the slope of the road, s be the distance of the road, a be the acceleration generated by Rob, 3a/4 is the acceleration generated by Jim .  Both of their motions are subjected to parallel component of the gravitational acceleration gsin\theta

Rob equation of motion can be modeled as s = a_Rt_R^2/2 = a300^2/2 = 45000a[/tex]

Jim equation of motion is s = a_Jt_J^2/2 = (3a/4)t_J^2/2 = 3at_J^2/8

As both of them cover the same distance

45000a = 3at_J^2/8

t_J^2 = 45000*8/3 = 120000

t_J = \sqrt{120000} = 346.4 s

So Jim should start 346.4 – 300 = 46.4 seconds earlier than Rob in other to reach the end at the same time

7 0
2 years ago
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho=rho0[1+α(T−T0)], where T0
Readme [11.4K]

Answer:

I = ΔVA[1 - α (T₀ - T)]/Lρ₀

Explanation:

We have the following data:

ΔV = Battery Terminal Voltage

I = Current through wire

L = Length of wire

A = Cross-sectional area of wire

T = Temperature of wire, when connected across battery

T₀ = Reference temperature

ρ = Resistivity of wire at temperature T

ρ₀ = Resistivity of wire at reference temperature

α = Temperature Coefficient of Resistance

From OHM'S LAW we know that;

ΔV = IR

I = ΔV/R

but,  R = ρL/A   (For Wire)

Therefore,

I = ΔV/(ρL/A)

I = ΔVA/ρL

but,   ρ = ρ₀[1 + α (T₀ - T)]

Therefore,

I = ΔVA/Lρ₀[1 + α (T₀ - T)]

I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹

using Binomial Theorem:

(1 +x)⁻¹ = 1 - x + x² - x³ + ...

In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).

Since, α generally has very low value. Thus, its higher powers can easily be neglected.

Therefore, using this Binomial Approximation, we can write:

[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]

Thus, the equation becomes:

<u>I = ΔVA[1 - α (T₀ - T)]/Lρ₀ </u>

3 0
2 years ago
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