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3 years ago

Desert sand is very hot in the day and very cool at night. what does this indicate about its specific heat capacity?

Physics

1 answer:

expeople1 [14]3 years ago

6 0

That the climate changes during the day and night high to low

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If Phoenix, Arizona, experiences a cool, wet day in June, does that mean the regions climate is changing?

jenyasd209 [6]

It doesn't mean that the climate is changing, it is probably just the morning dew.

5 0

3 years ago

Compare the benefits of wildfires to grasslands, northern forests, and deciduous forests.

Dimas [21]

Wildfires benefit grasslands, northern forests, and deciduous forests. Grasslands are benefited by improved soil quality and control of tree cover. Invertebrate species diversity is maintained through wildfire as well. Northern forests, like grasslands, experience increased production and nutritional quality of food as a result of wildfires. Deciduous forests experience an increase in the nutritional quality of food as well, but the effects are more temporary. The amount of shrubs in deciduous forests is reduced as a result of wildfires, allowing more herbaceous plants such as mosses and lichens to grow.

6 0

3 years ago

Read 2 more answers

……………………………………………………………..?

kotegsom [21]

Answer:

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

3 0

2 years ago

How does the magnitude of the electrical force between a pair of charged particles change when they are brought to half their or

Ostrovityanka [42]

Answer:

5. Quadruple

Explanation:

The electrostatic force between two charged particles is given by:

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

If the distance between the charges is reduced to half,

$r' = \frac{r}{2}$

So the new force will be

$F'=k\frac{q_1 q_2}{(r/2)^2}=4(k\frac{q_1 q_2}{r^2})=4F$

So, the force will quadruple.

4 0

3 years ago

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$

Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.

The mass and charge of a proton are:

m = 1.67 * 10^-27 kg

q = 1.6 * 10^-19 C

So, we get that the radius r will be:

r = $\frac{1.67 * 10^-27 kg * 2.2*10^6 m/s}{1.6 * 10^-19 C* 0.7 T}$ = 0.0328 m, or 3.28 cm.

8 0

3 years ago