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3 years ago

How might a location's altitude influence it's climate?

Physics

1 answer:

UNO [17]3 years ago

3 0

Answer:

You will discover that at high altitude, there is cold and the opposite is experienced when you go deep down the sea. However, the reason elevation affects climate and temperature gets colder is this. As you go higher up, the atmosphere experiences less pressure.

Explanation:

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Answer:

both answer is option C

Explanation:

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3 years ago

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3 years ago

Read 2 more answers

Consider the following three statements: For any electromagnetic radiation, the product of the wavelength and the frequency is a

boyakko [2]

Answer:

The correct answer is B. Statements (i) and (ii) are true.

Explanation:

Fisrt statement:

Electromagnetic radiation such as wave, wavelength λ and oscillation frequency ν are related by a constant, the speed of light. The equation is given by:

$c= \lambda \nu$

So the first statement is true.

Second statement:

$c=\lambda \nu\\c= 3.0 A \times 1.0\times 10^{18} Hz\\c= 3.0\times 10^{-10} m \times 1.0\times 10^{18} s^{-1}\\c=3\times 10^8 \frac{m}{s}$

The value of c in the vacuum is 3×10⁸ m/s. Hence, the second statement is true.

Third statement:

The speed of any electromagnetic radiation is constant regardless the type of radiation.

Hence, the third statement is false.

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3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$