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4vir4ik [10]
4 years ago
11

Calculate the power of an engine which can supply 60kg of energy per minute​

Physics
1 answer:
Nutka1998 [239]4 years ago
6 0

Answer:

1 Watt

Explanation:

Power (P) = w / Δt

w = work done

Δt = change in time

Power (P) = 60kg / 60 seconds

Power (P) = 1 Watt

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Julie is cycling at a speed of 3.4 meters/second. If the combined mass of the bicycle and Julie is 30 kilograms, what is the kin
nexus9112 [7]

Answer:

A

Explanation:

KE = 1/2 mv^2

=1/2(30kg)( 3.4 m/s)^2

=173.4 joules

=1.7×10^2 joules

6 0
3 years ago
Could someone summarize these into a paragraph? i'll give brainliest<br> DUE TODAY!!
slamgirl [31]

Answer:

I can't see the picture

Explanation:

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6 0
3 years ago
Read 2 more answers
Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of
EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

      E_{n}= k e² / 2a₀ (1 /n²)

      ao = h'² / k m e²               h' = h/2πi

For another atom with a single electron in the last layer

      a₀ ’= h’² / k m (Ze)²  

      a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

      E_{n} = - Z²  Eo/n²

     E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

         E_{n} -  E_{m} = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

       c = λ f

      E = h f

      E = h c /λ

      h c / λ = Z² ΔE

     λ = 1 / Z² (hc / ΔE)

     λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

     E_{n} = - 9 Eo / n²

     

      40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

      λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

     40.5 10-9 = 1/9 λ_hydrogen

    tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

      40.5 10⁻⁹ = 1/9 (16 \lambda_{Be} )

    tex]\lambda_{Be}[/tex] = 40.5 9/16

  tex]\lambda_{Be}[/tex] = 22.78 nm

6 0
3 years ago
You are trying to overhear a juicy conversation, but from your distance of 23.0 m , it sounds like only an average whisper of 40
Len [333]
If its asking the distance for the 65 db then use a proportion, if otherwise pleas clarify. It sounds like a pretty juicy conversation.
4 0
4 years ago
A coil of wire with 100. circular turns of radius 8.00 cm is in a uniform magnetic field along the axis of the coil. The resista
andreyandreev [35.5K]
I think Im gonna have to go with C 6.00 T/s but Im not sure
6 0
4 years ago
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