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olya-2409 [2.1K]
2 years ago
12

A toy helicopter has a mass of .25 kg. The rotors of the helicopter exert an upward lift

Physics
1 answer:
Paha777 [63]2 years ago
4 0
Nrhdndvdneg 5+5h hrndhsndhdks ndgsbdhdnd hendhhhhg ghetto hug the day the r the rhrbr end was the first day yyyyyyyyy I wanna do y’all have fun tonight and tomorrow morning and then
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Anything that floats in some fluid must be less dense than the fluid.
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PLEASE ANSWER ASAP!!
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As we know that magnetic field strength due to solenoid is given by

B = \frac{\mu_0 N i}{L}

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\mu_0 = permeability of air medium

N = number of turns

L = length of solenoid

i = electric current

Now as we enter the iron rod into the solenoid the medium will get change that will increase the permeability

so with the iron rod we will have

B = \frac{\mu_0 \mu_r N i}{L}

so correct answer will be

<u>It increases the magnetic field’s strength. </u>

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2 years ago
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3 years ago
A projectile launcher has a mass of 3 kg. It fires a projectile of mass 0.08 kg horizontally at a speed of 300 m/s. a. What is t
alexdok [17]

Answer:

0

Explanation:

It’s before the projectile was fired, so nothing has happened yet.

6 0
2 years ago
The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
  • the velocity of water in the upper channel is zero (v₁ = 0 m/s).
  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
  • g = 9.81 m/s²
  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

6 0
3 years ago
Read 2 more answers
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