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zalisa [80]
3 years ago
9

A 4113 N piano is to be pushed up a(n) 3.99 mfrictionless plank that makes an angle of 25.5â—¦with the horizontal.Calculate the

work done in sliding the piano
Physics
1 answer:
Serhud [2]3 years ago
3 0

Answer:

W = 14.8 kJ

Explanation:

W = F S cos ∅

W = 4113 x 3.99 x cos 25.5

W = 16410.87 x 0.9025 = 14810.8 J or 14.8 kJ

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Technician A says that current is created in the stator windings of an alternator. Technician B says that the voltage regulator
drek231 [11]

Answer:

Technician A and Technician B both are right.

Explanation:

In an AC alternator,  there are two windings

1. Stator winding (stationary)

2. Rotor winding (rotating)

The current is induced in the stationary coils due to the magnetic field produced by the rotor. The DC suppy is provided to the rotor winding via slip rings and brushes and a voltage regulator precisely controls this supply to control the current flow through the rotor.

Therefore, both technicians are right.

4 0
3 years ago
What is the Orbital Notation for Radon
Marina CMI [18]

Answer:

[Rn]7s2

Explanation:

3 0
3 years ago
One of the difficulties in studying mechanics is that many common words are used with highly specific technical meanings,among t
Allisa [31]

Mnanan anaw anwkkww wjakana akkqna akaoaman akamakq

8 0
3 years ago
Energy from the Sun arrives at the top of the Earth's atmosphere with an intensity of 1.36 kW/m 2 . How long does it take for 3.
alina1380 [7]

Answer:

The time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

Explanation:

Given the intensity of the energy is 1.36\ kW/m^2

And the arriving energy is 3.55\times 10^9\ J

Also, the area in which energy is being arriving is 4.25\ m^2

Now, we will use relation between energy (E), intensity of energy (p), area (A) and time (T).

Where energy is in Joule, intensity is in kW/m^2, area is in m^2 and time is in seconds.

The equation is

E=pAT\\\\T=\frac{E}{pA}\\\\T=\frac{3.55\times 10^9}{1.36\times 10^3\times 4.25}\\\\T=0.614\times 10^6\ s\\T=6.14\times 10^5\ seconds

So, the time taken by 3.55\times 10^9\ J to arrive on an area of 4.25\ m^2 is 6.14\times 10^5\ seconds.

4 0
3 years ago
When cs-137 decays, it emits gamma radiation. the energy of one photon is 1.06 × 10-13 j. what is the wavelength of this radiati
Kryger [21]
To find the solution to the problem, we would be using Planck's equation which is E = hv
Where:
E = energy
h = Planck's constant = 6.626 x 10-34 J·s
ν = frequency
Then, you’ll need a second equation which is c = λν
Where:
c = speed of light = 3 x 108 m/sec
λ = wavelength
ν = frequency
Reorder the equation to solve for frequency:ν = c/λ
Next, substitute frequency in the first equation with c/λ to get a formula you can use:
E = hν
E = hc/λ
But we are looking for the wavelength, so rearrange it more, then our final equation would be:
λ = hc / E 
λ = (6.625E-34)(3.0E8 m/s) / (1.06E-13) 
λ = 1.875E-12 m
3 0
3 years ago
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