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galina1969 [7]
2 years ago
11

Write two examples of machine which changes the direction of force​

Physics
1 answer:
Ostrovityanka [42]2 years ago
4 0

Answer:

Washign machine and swingset

Explanation:

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What does the equation n =Pout/Pin mean
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<span>Basically, the variable n is equal to the amount of P out of the system divided by the about of P into the system

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3 years ago
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Un alambre de teléfono de 120m de largo y de 2.2mm de diámetro se estira debido a una fuerza de 380 N cual es el esfuerzo longit
sergij07 [2.7K]

Respuesta: verifique amablemente la explicación

Explicación:

Dado lo siguiente:

Longitud (L) del cable = 120 m

Diámetro (d) = 2,2 mm (2,2 / 1000) = 2,2 * 10 ^ -3 m

Fuerza (F) = 380 N

Esfuerzo longitudinal = Fuerza / Área

Área = πd² / 4 = (π * (2.2 * 10 ^ -3) ^ 2) / 4

Área = (3.142 * 4.84 * 10 ^ -6)

Área = 0.00000380132 m²

Estrés = Fuerza / Área

Estrés = 380 / 0.00000380132

Esfuerzo longitudinal = 99952128.12 = 9.9952128 * 10^7 Nm^-2

Deformación longitudinal: extensión / longitud

Extensión = 0.10 m

Longitud = 120 m

Deformación longitudinal = 0,1 m / 120 m

Deformación longitudinal = 0.0008333 = 8.33 × 10 ^ -4

6 0
2 years ago
2 (a) A plane from airport A flies 280 km to the east to airport B. The plane then travelled north to airport C, 190 km away. (i
Mekhanik [1.2K]

Answer:

See below

Explanation:

See attached diagram

280 km   east then 190 km north

Use Pythagorean theorem to find the resultant displacement

d^2 = 280^2 + 190^2

d = 338.4 km

Angle will be    arctan ( 190/280) = 34.16 °

4 0
1 year ago
The<br> is the particle in the nucleus with a positive charge.<br> Answer here
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Answer: A proton is a positively charged particle located in the nucleus of an atom.
3 0
3 years ago
A spherical gas-storage tank with an inside diameter of 9 m is being constructed to store gas under an internal pressure of 1.50
lutik1710 [3]

Answer: 33 mm

Explanation:

Given

Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m

Internal pressure of gas, P(i) = 1.5 MPa

Yield strength of steel, P(y) = 340 MPa

Factor of safety = 0.3

Allowable stress = 340 * 0.3 = 102 MPa

σ = pr / 2t, where

σ = allowable stress

p = internal pressure

r = radius of the tank

t = minimum wall thickness

t = pr / 2σ

t = 1.5*10^6 * 4.5 / 2 * 102*10^6

t = 0.033 m

t = 33 mm

The minimum thickness of the wall required is therefore, 33 mm

6 0
3 years ago
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