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Mama L [17]
3 years ago
14

How lift force works in CER format

Physics
1 answer:
Step2247 [10]3 years ago
3 0
The really good answer is 30cm CER format
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Air flows upward in the wick of a lantern because of the liquid property called
vovangra [49]

Adhesive.

Adhesive is the force of attraction between molecules of different kind. Liquid flows upward the wick because the adhesive force between the wick and the liquid is higher than cohesive forces in the liquid.

When the adhesive  force  between the wick and the liquid is high we have capillarity taking place. This cause the liquid to move up the wick.

7 0
3 years ago
You are standing a distance of 17.0 meters from the center of a merry-go-round. The merry-go-round takes 9.50 seconds to go comp
GenaCL600 [577]

Answer:

(a)11.24 m/s

(b)7.44 m/s

(c)409 N

(d)539.55\mu

(e) 0

Explanation:

The period for 1 circle 2\pi of the merry go around is 9.5s. It means the angular speed is:

\omega = \theta / t = 2\pi / 9.5 \approx 0.661 rad/s

(a)The speed is

v = \omega * R = 0.661 * 17 = 11.24 m/s

(b) Centripetal acceleration:

a = \frac{v^2}{R} = \frac{11.24^2}{17} = 7.44 m/s^2

(c) Magnitude of the force that keeps you go around at this acceleration

F = ma = 55 * 7.44 = 409 N

(d) let the coefficient of friction by \mu. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

F_f = mg\mu = 55*9.81\mu = 539.55\mu

3 0
3 years ago
1. A limiting factor for using nuclear energy is the
Goryan [66]

Answer:

waste

Explanation:

took da test. nhgvgg

8 0
3 years ago
Read 2 more answers
a cylinder has a radius of 2.1cm and a length of 8.8cm .total charge 6.1 x 10^-7C is distributed uniformly throughout. the volum
7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 ×10^{-2} m

height (h) =8.8 cm = 8.8 ×10^{-2} m

total charge (q) =6.1×10^{-7} C

To solve:

volume charge density (ρ)

We know;

<u> ρ =q ÷ v</u>

volume of cylinder = π ×r  × r × h

volume of cylinder =3.14 × 2.1 × 2.1 ×10^{-4} × 8.8 ×10^{-2}

volume of cylinder (v) = 122.23 ×10^{-6}

<u> ρ =q ÷ v</u>

ρ = 6.1×10^{-7} ÷   122.23 ×10^{-6}

<u>ρ = 0.04 × </u>10^{-1}<u> C</u>

Therefore the volume charge density of a substance (ρ) is 0.04 × 10^{-1} C.

3 0
3 years ago
Water enters a typical garden hose of diameter 0.016 m with a velocity of 3 m/s. Calculate the exit velocity of water from the g
Vladimir79 [104]

Answer:

v₂ = 306.12 m/s

Explanation:

We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:

A₁v₁ = A₂v₂

where,

A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²

v₁ = entrance velocity = 3 m/s

A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²

v₂ = exit velocity = ?

Therefore,

(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂

v₂ = (0.006 m³/s)/(0.0000196 m²)

<u>v₂ = 306.12 m/s</u>

5 0
3 years ago
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