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ser-zykov [4K]
4 years ago
15

Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun

ny day, you notice that the air close to the ground feels warmer. which process explains this?
Physics
1 answer:
yanalaym [24]4 years ago
6 0
The answer is convection
You might be interested in
PLEASE HELP!! At what temperature will silver have a resistivity that is four times the resistivity of tungsten at room temperat
Svetach [21]
Consider 20 deg.C. as room temperature.

From tables,
Silver has a resistivity of  1.6*10^-8 ohm-m at 20 deg.C, and it increases by 0.0038 ohm-m per deg.K increase.
Therefore if the temperature rise above 20 deg.C is T, then silver will have resistivity of
1.6*10^-8(1 + 0.0038T) ohm-m

At room temperature, the resistivity of tungsten (from tables) is 5.6*10^-8.

The resistivity of silver will be 4 times that of tungsten (at room temperature) when
1.6*10^-8(1 + 0.0038T) = 4*5.6*10^-8
1 + 0.0038T = 14
T = 13/.0038 = 3421 deg.K approx

Answer: 20 + 3421 = 3441 °C
4 0
3 years ago
During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m
Nezavi [6.7K]

Answer:

<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

Explanation:

Given :

Current I = 2.5 \times 10^{4} A

Time t = 40 \times 10^{-6} sec

We know that the current is the rate of flow of charge.

From the formula of current,

<h3>  I = \frac{Q}{t}</h3>

Where Q = charge transfer between cloud and earth.

 Q =I t

Q = 2.5 \times 10^{4} \times 40 \times 10^{-6}

Q = 1 C

Hence, the charge transferred from the cloud to earth is 1 Coulomb.

3 0
4 years ago
If you given volume. for example 200 cm³ how can you change it to area m²​
PIT_PIT [208]

Answer:

move the decimal 6 places to the left.

Explanation:

um I assume you meant to say area m^3

7 0
3 years ago
Pls help need it NOW
kaheart [24]
It’s a... in the nucleus
4 0
3 years ago
Read 2 more answers
A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

3 0
3 years ago
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