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Kamila [148]
4 years ago
10

A piano with a mass of 130 kg is lifted 10m above the ground in 5 s by a crane.

Physics
2 answers:
hammer [34]4 years ago
7 0
The 2548 watts should be the power that is used by the crane according to the formula
Sergio [31]4 years ago
6 0

<em>power = 2548 \: watt \\ solution \\ mass = 130 \: kg \\ distance = 10m \\ time = 5 \: sec \\ power =  \frac{w}{t}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{f \times d}{t}  \\  \:  \:  \:  \:  \:  \:  \:  \:  =  \:  \frac{m \times g \times d}{t}  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{130 \times 9.8  \times 10}{5}  \\  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{12740}{5}  \\  \:  \:  \:  \:  \:  = 2548 \: watt \\ hope \: it \: helps</em>

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1 x 10^-9 N

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3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

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c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

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hence it is a subsonic aircraft

4 0
3 years ago
An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag
Afina-wow [57]

Answer:

The induced emf in the loop is 7.35\times 10^{-4}\ V

Explanation:

Given that,

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4a = 1.22

a = 0.305 m

Area of square, A=a^2=(0.305)^2=0.0930\ m^2

Circumference of the loop,

C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m

Area of circle,

A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2

The induced emf is given by :

\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V

So, the induced emf in the loop is 7.35\times 10^{-4}\ V

8 0
3 years ago
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