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ElenaW [278]
3 years ago
13

What causes turbulence?

Physics
1 answer:
melomori [17]3 years ago
6 0

Answer:

B

Explanation:

I think

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zysi [14]

Answer:

education

clothes

food

shelter

Explanation:

7 0
3 years ago
Read 2 more answers
A 60-kg uniform board 2.4 m long is supported by a pivot 80 cm from the left end and by a scale at the right end (see the Fig. b
yKpoI14uk [10]

Answer:

a) x = 0.61 m

b) x = 1.4 m

Explanation:

Given data;

M = 60 kg,

m = 40 kg,

F considered to be a  reading

Torque of N about the given pivot point = F*(2.4 - 0.80) = 1.6 F (counterclockwise)

Torque of Mg about the given pivot Point = Mg*(1.2 - 0.80)

= 0.4*60*9.8 = 235.2 N-m (clockwise)

a) F = 100 N

note 1.6F = 160 < 235.2, so

mass m should be placed at left of pivot,

its torque = mg*(0.80 - x)

                = 40*9.8*(0.80 - x) = 392(0.80 -x)

160 + 392(0.8 - x) = 235.2

x = 0.61 m

b) N = 300 N

note 1.6F = 480 > 235.2, so

mass m should be placed at right of pivot,

its torque = mg*(x - 0.80) = 40*9.8*(x - 0.80) = 392(x -0.80)

480 = 392(x - 0.8) + 235.2

x = 1.4 m

3 0
3 years ago
Suppose that she pushes on the sphere tangent to its surface with a steady force of F = 60 N and that the pressured water provid
sergey [27]

Answer:

25.06s

Explanation:

Remaining part of the question.

(A large stone sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water.)

Solution:

F = 60N

r = 90cm = 0.9m

M = 8200kg

Moment of inertia for a sphere (I) = ⅖mr²

I = ⅖ * m * r²

I = ⅖ * 8200 * (0.9)²

I = 0.4 * 8200 * 0.81

I = 2656.8 kgm²

Torque (T) = Iα

but T = Fr

Equating both equations,

Iα = Fr

α = Fr / I

α = (60 * 0.9) / 2656.8

α = 0.020rad/s²

The time it will take her to rotate the sphere,

Θ = w₀t + ½αt²

Angular displacement for one revolution is 2Π rads..

θ = 2π rads

2π = 0 + ½ * 0.02 * t²

(w₀ is equal to zero since sphere is at rest)

2π = ½ * 0.02 * t²

6.284 = 0.01 t²

t² =6.284 / 0.01

t² = 628.4

t = √(628.4)

t = 25.06s

8 0
4 years ago
Suppose that the ultrasound source placed on the mother's abdomen produces sound at a frequency 2 MHz (a megahertz is 10^610 ​6
ella [17]

Answer:

the maximum frequency observed is 2.0044 10⁶ Hz

Explanation:

This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is

          f ’= f_o \ ( \frac{v \pm  v_o}{v} )

the + sign is used when the observer approaches the source

typical speeds of a baby's heart stop are around 200 m / min

let's reduce to SI units

        v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s

let's calculate

         f ’= 2 10⁶ (\frac{1500 \ \pm 3.33}{1500})  

         f ’= 2.0044 10⁶ Hz

         f ’= 1,9956 10⁶ Hz

therefore the maximum frequency observed is 2.0044 10⁶ Hz

8 0
3 years ago
The volume of a gas can be converted to moles by
Sonja [21]
Multiplying the ideal gas law constant
7 0
3 years ago
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