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Hoochie [10]
3 years ago
5

Based on how solar radiation received varies with latitude, it can be inferred that seasonal temperature contrasts would _______

_ as latitude increases
Physics
2 answers:
Andrei [34K]3 years ago
4 0

Answer:

Increase

Explanation:

As the latitude increases, the angle at which the Sun light falls on the Earth changes. This is known as the Sun angle. Although the rays fall on the Earth parallely but they make different angles with objects at different latitudes due to Earth's curvature.

We know that the throughout the year the Sun moves in the sky and changes its altitude but it never gets overhead for a latitude above or below the tropics. Due to this phenomenon, the Sun angle doesn't vary much for tropical region whereas it varies a lot for the region beyond tropics. If the Sun angle varies too much the contrast between the seasonal temperature also increases.

For example: the Sun angle for equator lies in the range of 0° - 23.5° while at the arctic circle (66.5° N) the range is 43° - 90°.

Klio2033 [76]3 years ago
3 0
The answer that best completes the statement above is the word INCREASE. Therefore, the <span>seasonal temperature contrasts would INCREASE as latitude increases. The existence of varied seasonal temperature is dependent on the latitude on earth depending on the location. Hope this answers your question.</span>
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Answer:

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Explanation:

Given that,

Mass, m = 27 grams

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d=\dfrac{m}{V}\\\\d=\dfrac{27}{15}\\\\d=1.8\ g/cm^3

So, the density of the substance is equal to 1.8\ g/cm^3.

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The electric potential at a position located a distance of 20.7 mm from a positive point charge of 8.60×10-9C and 15.1 mm from a
max2010maxim [7]

Answer:

q2 = -4.35*10^-9C

Explanation:

In order to find the values of the second charge, you use the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}       (1)

V: electric potential = 1.14 kV = 1.14*10^3 kV

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1: charge 1 = 8.60*10^-9 C

q2: charge 2 = ?

r1: distance to the first charge = 20.7mm = 20.7*10^-3 m

r2: distance to the second charge = 15.1mm

You solve the equation (1) for q2, and replace the values of the other parameters:

q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C

The values of the second charge is -4.35*10^-9C

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1. There is a famous intersection in Kuala Lumpur, Malaysia, where thousands of vehicles pass each hour. A 750 kg Tesla Model S
tigry1 [53]

Solution :

Let the positive x-axis is along the East and the positive y direction is along the north.

Given :

Mass of the Tesla car, m_1 = 750 \ kg

Mass of the Ford car, m_2 = 1250 \ kg

Now let the initial velocity of Tesla car in the south direction be = -v_1j

The initial momentum of Tesla car, p_1 = -750 \ v_1

Let the initial velocity of Ford car in the east direction be = v_2 \ i

So the initial momentum of the Ford car is p_2=1250\ v_2 \ i

Therefore, the initial velocity of both the cars is p_i = p_1+p_2

                                                                  =1250 \ v_2 \ i - 750\ v_1 \ j

Now the final velocity of both the cars is v = 18 \ m/s

So the vector form is :

v = 18\cos 32\ i-18 \sin 32 \ j

  = 15.26 \ i - 9.54 \ j

Therefore the momentum after the accident is

p_f=(m_1+m_2) \times v

    =(750+1250) \times (15.26 \ i - 9.54 \ j)

    = 30520\ i -19080\ j

According to the law of conservation of momentum, we know

p_i = p_f

1250 \ v_2 \ i - 750\ v_1 \ j  = 30520\ i -19080\ j

1250 \ v_2 = 30520

v_2=24.4 \ m/s

From, 750\ v_1 = 19080

We get, v_1=25.4 \ m/s

Therefore the speed of Tesla car before collision = 25.4 m/s

The speed of ford car before collision = 24.4 m/s

6 0
3 years ago
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