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vlabodo [156]
3 years ago
6

Why do surface winds tend to cross the isobars?

Physics
2 answers:
Rudik [331]3 years ago
6 0
Geostrophic wind<span> blows parallel to the </span>isobars<span> because the Coriolis force and pressure gradient force are in balance. However it should be realized that the actual </span>wind<span> is not always geostrophic. Especially near the </span>surface<span>. The </span>surface<span> of the Earth exerts a frictional drag on the air blowing just above it.</span>
scZoUnD [109]3 years ago
3 0
Geostrophic winds blows parallel to the isobars. That is because the Coriolis force and pressure gradient force ( PGF ) are in balance. But near the surface the friction can act to change the direction of the wind and to slow it down. Coriolis force decreases at the surface and PGF stays the same. The difference in terrain conditions affects how much friction is exerted. Hills and forests force the wind to change direction more than flat areas. Answer: Friction reduces the speed so Coriolis is weakened. 
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Timbre describes the sound’s _____.<br><br> pitch<br><br> loudness<br><br> tone<br><br> quality
Bumek [7]

Answer:

A) pitch

Explanation:

I did band and also g*ogled it to check :)

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7 0
2 years ago
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A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C an
nekit [7.7K]

Answer:

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV................ Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2............. Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

3 0
3 years ago
What are six countries that are islands near Florida​
vodomira [7]

Answer:

Cuba, Jamaica, Dominican Republic, Haiti, Turks & Caicos, and The Bahamas

Explanation:

4 0
3 years ago
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You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity le
BigorU [14]

Answer:

1) the new power coming from the amplifier is 19.02 W

2) The distance away from the amplifier now is 5.50 m

3) u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

Explanation:

Lets say that I am at a distance "u" from the TV,

Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB

SO

S(indB) = 10log (I₁/1₀)

we substitute

125 = 10(I₁/10⁻¹²)

12.5 = log (I₁/10⁻¹²)

10^12.5 = I₁/10^-12

I₁ = 10^12.5 × 10^-12

I₁ = 10^0.5 W/m²

Now I₂ will be intensity of sound when corresponding sound level is 107 dB

107 = 10log(I₂/10⁻²)

10.7 = log(I₂/10⁻¹²)

10^10.7 = I₂ / 10^-12

I₂ = 10^10.7  ×  10^-12

I₂ = 10^-1.3 W/m²

Now since we know that

I = P/4πu² ⇒ p = 4πu²I

THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂

Therefore

P₁/P₂ = I₁/I₂

WE substitute

P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)

P₂ = 19.02 W

the new power coming from the amplifier is 19.02 W

2)

P₁ = 4πu²I₁

u =√(p₁/4πI₁)

u = √(1200/4π × 10^0.5)

u = 5.50 m

The distance away from the amplifier now is 5.50 m

3)

Let I₃ be the intensity corresponding to required sound level 85 dB

85 = 10log(I₃/10⁻¹²)

8.5 = log (I₃/10⁻¹²)

10^8.5 = I₃ / 10^-12

I₃ = 10^8.5  × 10^-12

I₃ = 10^-3.5 w/m²

Now, I ∝ 1/u²

so I₂/I₃ = u₁²/u²

u₁ = √(I₂/I₃) × u

u₁ = √(10^-1.3 / 10^-3.5) ×  5.50

u₁ = 69.24 m

Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther

8 0
3 years ago
In the forest (phrase—noun, but no subject and no verb) Smashing against the bat (phrase—verb but no subject) Before our first c
gayaneshka [121]

Yes all those are correct, but I don't know what the question was

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