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3 years ago

The amount of water vapor in the air is known as the _______.

Physics

2 answers:

storchak [24]3 years ago

5 0

It is C the humidity

Bond [772]3 years ago

5 0

<span>The amount of water vapor in the air is known as the _______.

c. Humidity

</span>

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PLS ANSWER FAST WILL GIVE BRAINLEST!!!

Kipish [7]

Answer:

2 Newtons

Explanation:

$F = ma$

Therefore, your mass would be 1kg and your acceleration would be 2m/s/s

Plug the numbers into the equation:

$(1kg)(2m/s/s)$

which will equal

$2 Newtons$

8 0

3 years ago

Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

Norma-Jean [14]

Answer:

Part a)

$\lambda = 300 m$

Part b)

$E = 2.7 N/C$

Part c)

$I = 9.68 \times 10^{-3} W/m^2$

$P = 3.22 \times 10^{-11} N/m^2$

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

$\lambda = \frac{c}{f}$

$\lambda = \frac{3\times 10^8}{1\times 10^6}$

$\lambda = 300 m$

Part b)

As we know the relation between electric field and magnetic field

$E = Bc$

$E = (9 \times 10^{-9})(3\times 10^8)$

$E = 2.7 N/C$

Part c)

Intensity of wave is given as

$I = \frac{1}{2}\epsilon_0E^2c$

$I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)$

$I = 9.68 \times 10^{-3} W/m^2$

Pressure is defined as ratio of intensity and speed

$P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}$

$P = 3.22 \times 10^{-11} N/m^2$

6 0

3 years ago

Which takes longer to reach earth, light from the sun or light from other stars? Explain.

densk [106]

Light from other stars take longer to reach the earth because they are farther than our sun.

4 0

3 years ago

Read 2 more answers

A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha

Nikolay [14]

Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

Radius of curve = 50 m

θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

$N = \dfrac{m g}{cos\theta-\mu sin \theta}$

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

$\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}$

taking cos θ from nominator and denominator

$F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg$

$\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g$

$\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}$

now, inserting all the given values

$\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}$

μ = 0.6

7 0

3 years ago

The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive towa

uysha [10]

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

vo = 15.5 m/s

6 0

2 years ago