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2 years ago

Desert climates can have very hot days and very cold nights. This is because _____.

2 answers:

7 0

There is usually a clear skhy

it provides a clear shot for the heat of the sun on the sand and sand can cool very quickly so it allows the air to cool just as fast

hope i helped
Dosvedanya :)

daser333 [38]2 years ago

6 0

Answer:

The correct answer is option B, Clear Sky

Explanation:

The temperature during the day time rises to maximum as there is no moisture in the atmosphere. The moisture being extremely low leads to formation of no clouds/or very rare clouds . Due to no clouds, sky is clear and thus the heat from the sun does not stay close to the Earth surface and it is radiated back to the maximum extent causing the temperatures at night falling to nearly freezing.

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Keisha is making a diagram of a simple machine in the body.

Amiraneli [1.4K]

Answer:

Sorry for being late. It is...

A.) X: Load, Y: Fulcrum, Z: Lever

3 0

2 years ago

Read 2 more answers

Part A

geniusboy [140]

Answer:

a) b = -5

b) slope = 3/2

Explanation:

a) The equation of a line is given as y = mx + b, where m is the slope of the line and b is the intercept on the y axis.

Given that y = 3x + b and it passes through the point (2, 1). Hence when x = 2, y = 1. Therefore, substituting for x and y:

1 = 3(2) + b

1 = 6 + b

b = 1 - 6

b = -5

b) The equation of a line passing through two points ( $x_1,y_1$ ) and $x_2,y_2$ is given by:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

The equation of the line passing through the two points (0,3) and (4,9) is:

$y-3=\frac{9-3}{4-0}(x-0)\\ \\y-3=\frac{3}{2}x\\ \\y = \frac{3}{2}x+3$

Comparing y = (3/2)x + 3 with y = mx + b, the slope (m) is 3/2

4 0

2 years ago

An AM radio transmitter broadcasts 63.2 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike

egoroff_w [7]

Answer:

Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

Explanation:

Given;

power of radio transmitter, P = 63.2 kW = 63200 W

distance of transmission, r = 30.5 km

Intensity of the transmitted radio wave is calculated as follows;

$I = \frac{P}{4\pi r^2}$

where;

I is the intensity of the transmitted radio wave

Substitute the given values and calculate the intensity of the transmitted radio wave;

$I = \frac{P}{4\pi r^2} = \frac{63200}{4\pi (30500)^2} = 5.406 *10^{-6} \ W/m^2$

Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

6 0

2 years ago

An infrared spectrometer on Dawn found something unexpected on Ceres's surface. Its presence suggested that Ceres might have for

Sati [7]

The question is missing alternatives. Here is the complete question.

An infrared spectrometer on Dawn found something unexpected on Ceres's surface. Its presence suggested that Ceres might have formed farther from the Sun, or been impacted by objects from a more-distant part of the solar system. What was this finding?

1. The fact that Ceres is covered with small dark particles that appear identical to the composition of Uranus's rings.

2. The presence of a thick cloud layer made of sulfuric acid, similar to what is observed at Venus.

3. The presence of clay-like minerals with ammonia bound up in them.

4. The infrared spectrum of Ceres's surface is essentially identical to that of most objects in the Kuiper Belt.

Answer: 3. The presence of clay-like minerals with ammonia bound up in them.

Explanation: The discovery of ammonia clay-like minerals in Ceres is surprising because it would be encoutered in planets that are far from the Sun, since ammonia requires colder temperatures, which is found beyond Jupiter's orbit, to condense. This finding can ascertain not only the origins of the dwarf planet as how the solar system was formed, were organized and evolved, because understanding where smaller planets are formed is important to determine their destiny.

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2 years ago

A small artery has a length of 1.10 × 10-3 m and a radius of 2.50 × 10-5 m. If the pressure drop across the artery is 1.15 kPa,

levacccp [35]

Answer:

$7.69533\times 10^{-11}\ m^3/s$

Explanation:

P = Pressure difference = 1.15 kPa

r = Radius = $2.5\times 10^{-5}\ m$

$\eta$ = Viscosity of liquid = $2.084\times 10^{-3}\ Pas$

l = Length of artery = $1.1\times 10^{-3}\ m$

From Poiseuille's equation we have

$Q=\frac{\pi Pr^4}{8\eta l}\\\Rightarrow Q=\frac{\pi 1.15\times 10^3\times (2.5\times 10^{-5})^4}{8\times 2.084\times 10^{-3}\times 1.1\times 10^{-3}}\\\Rightarrow Q=7.69533\times 10^{-11}\ m^3/s$

The flow rate of blood is $7.69533\times 10^{-11}\ m^3/s$

8 0

2 years ago