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Furkat [3]
3 years ago
14

What is the name of the device that measures wind speed?

Physics
2 answers:
myrzilka [38]3 years ago
8 0
The device is an anemometer<span> 
It's like 4 cups that measure the speed instead of direction

</span>
Igoryamba3 years ago
3 0
"Anemometer" is the device that measures wind speed

Hope this helps!
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What is the potential energy of an object 20 m in the air with a<br> mass of 600 kg?
Lana71 [14]

Answer:

Ep = 117600 J

Explanation:

Data:

  • Mass (m) = 600 kg
  • Height (h) = 20 m
  • Gravity (g) = 9.8 m/s²
  • Potential Energy (Ep) = ?

Use formula:

  • Ep = m * g * h

Replace:

  • Ep = 600 kg * 9.8 m/s² * 20 m

Multiply operations, and units:

  • Ep = 117600 J

What is the potential energy?

The potential energy is <u>117600 Joules.</u>

7 0
2 years ago
Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw
aalyn [17]
Ok, I think this is right but I am not sure:
 Q = ϵ
 0AE
A= π π
 r^2

=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons


4 0
3 years ago
the force exerted by gravity,electricity,and magnetism act at a distance with no physical connection. for this reason, they are
Soloha48 [4]
Its called the electric force
7 0
3 years ago
Read 2 more answers
Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu
Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega

And so, the current through the circuit is (using Ohm's law):

I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

P=I^2 R

where I is the current and R is the resistance.

For the first bulb:

P_1 = (0.1 A)^2 (400 \Omega)=4 W

For the second bulb:

P_1 = (0.1 A)^2 (800 \Omega)=8 W

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W

3 0
3 years ago
If
Aleksandr-060686 [28]

Answer:

A factor of 2*4 = 8

Explanation:

F_g = (G*m1*m2)/r^2  

where m1 and m2 are the two masses, G is Newton's gravitational constant, and r is the distance between the center of mass of the two objects.

So, if you double m1 and quadruple m2:

m1' = 2*m1

m2' = 4*m2

Then F_g' = (G*m1'*m2')/r^2 = (G*2*m1*4*m2)/r^2 = 8*(G*m1*m2)/r^2 = 8*F_g

3 0
2 years ago
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