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tamaranim1 [39]
3 years ago
11

Einstein is probably most known for his equation E = mc2, where E = the energy of a particle of mass = m, and c = the speed of l

ight. If "m" and "c" are in MKS units, what are the MKS units of E?
Physics
1 answer:
Zepler [3.9K]3 years ago
6 0

Answer:

speed of light

Explanation:

it is speed because they are talking about speed in the equation

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A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?
Olenka [21]
To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
                             F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N. 
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3 years ago
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Find the impedance of the load for maximum power transfer in the attached circuit.
Tanya [424]
I can't find the attached circuit.

Maximum power transfer occurs when the load impedance
is equal to the internal impedance of the power source.
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3 years ago
This is how sodium appears in the periodic table. An orange box has N a at the center and 11 above. Below it says sodium and bel
Vera_Pavlovna [14]

Answer:

1 valence electrons

11 prontons

12 neutrons

Explanation:

N_{11} = 1s^{2} 2s^{2} 2p^{6} 3s^{1} \\

1 valence electrons

p = 11

11 prontons

A = p + n

A = 22,99 = 23

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You have just landed your first job as a structural engineer and you have been asked to design a regional airport for small plan
lidiya [134]

Answer: 197.40\ m

Explanation:

Given

final velocity at takeoff v=28.1\ m/s

Acceleration of the plane can be a=2\ m/s^2

Initial velocity is zero for the plane i.e. u=0

Using the equation of motion

\Rightarrow v^2-u^2=2as\quad [\text{s=displacement}]\\\text{Insert the values}\\\Rightarrow (28.1)^2-0=2\times 2\times s\\\\\Rightarrow s=\dfrac{789.61}{4}\\\\\Rightarrow s=197.40\ m

Thu,s the minimum length must be 197.40\ m

6 0
2 years ago
a 58 kg sprinter, starting from rest, runs a 60 m long race in 7.8 s with a constant acceleration. how much work does the sprint
Alex777 [14]

Work done by the sprinter in first 2.5 s will be 702.699 J.

Work done is the product of force and distance covered.

  • Mathematically, W = Force * Distance

Mass of the sprinter = 58 kg

Distance covered by the sprinter = s = 60 m

Total time taken by the sprinter = t = 7.8 s

  • According to the Equations of motion S = ut + \frac{1}{2}at^{2}

u = initial velocity = 0

a = constant acceleration

60 =  \frac{1}{2}a7.8^{2}

60.84 a = 60 * 2

a = 1.97 m/s^{2}

For a = 1.97 m/s^{2} and t = 2.5 s

d =  \frac{1}{2}at^{2}

Here also u is '0' because the distance is being measured from the starting point.

d = \frac{1}{2} *1.97*(2.5)^{2}

d = 6.15 m

Work done by the sprinter = F * d

Force = mass * acceleration

F = 58 * 1.97

F = 114.26 N

distance for first 2.5 sec is 6.15 m

Therefore Work done = 6.15 * 114.26

W = 702.699 J

Work done by the sprinter is 702.699 J.

To know more about Work done,

brainly.com/question/28338228

#SPJ4

8 0
10 months ago
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