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3 years ago

At 45 degrees s latitude, the angle of the noon sun is lowest and the length of daylight is shortest on:

Physics

1 answer:

Marat540 [252]3 years ago

8 0

Answer;

- June 21

At 45 degrees latitude, the angle of the noon sun is lowest and the length of daylight is shortest on June 21.

Explanation;

-On June 21 you will note that the Northern Hemisphere is pitched toward the Sun. This means that the Sun's vertical ray is striking the Earth at the Tropic of Cancer (23.5 degrees N).

-Days tend to get longer in the northern hemisphere from December 21 to June 21, and then grow shorter from June 21 to December 21. The June solstice is the summer solstice in the Northern Hemisphere and the winter solstice in the Southern Hemisphere. The date varies between June 20 and June 22, depending on the year, and the local time zone.

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Sedbober [7]

Answer:

Explanation:

The water collects in the ocean; it is then evaporated by the sun. After evaporation the water turns into water vapor, it then condenses to form clouds.

3 0

2 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

2 years ago

What would you do to increase resistance

ratelena [41]

Answer:

If this is electrical currents , make the wire longer, smaller diameter wires, heat it up

6 0

2 years ago

Read 2 more answers

Some superconductors are capable of carrying a very large quantity of current. If the measured current is 1.00 ´ 105 A, how many

Zielflug [23.3K]

Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

Charge of electron $e = 1.6 \times 10^{-19}$ C

Time $t = 1$ sec

From the formula of current,

Current is the number of charges flowing per unit time.

$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

5 0

3 years ago

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

$E= \Delta mc^{2}$

where $\Delta m$ = change in mass

c = speed of light = $3 \times 10 ^{8}m/s$

Making m the subject of the formula, we can find the change in mass to be

$\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg$

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

5 0

3 years ago