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Setler79 [48]

Answer:

Speed of light

Explanation:

The famous Einstein's equation is:

$E=mc^2$

where

E is the energy

m is the mass

$c=3\cdot 10^8 m/s$ is the speed of light

In this equation, Einstein summarized the following fact: mass can be converted into energy, and the amount of energy released in such a process is given by the equation.

An example of application of this equation is the nuclear fusion process. In a nuclear fusion, two lighter nuclei combine into a heavier nucleus. However, the mass of the heavier nucleus is slightly less than the sum of the masses of the two original nuclei: some of the mass of the original nuclei has been converted into energy, accorging to the previous equation.

4 0

3 years ago

Please help me find the equations guys

Alexeev081 [22]

The line at the bottom of the picture ... probably the first line on a list of choices .. is the correct equation.

4 0

3 years ago

Sam receives the kicked football on the 3 yd line and runs straight ahead toward the goal line before cutting to the right at th

Pie

Answer:

Distance: 21 yd, displacement: 15 yd, gain in the play: 12 yd

Explanation:

The distance travelled by Sam is just the sum of the length of each part of Sam's motion, regardless of the direction. Initially, Sam run from the 3 yd line to the 15 yd line, so (15-3)=12 yd. Then, he run also 9 yd to the right. Therefore, the total distance is

d = 12 + 9 = 21 yd

The displacement instead is a vector connecting the starting point with the final point of the motion. Sam run first 12 yd straight ahead and then 9 yd to the right; these two motions are perpendicular to each other, so we can find the displacement simply by using Pythagorean's theorem:

$d=\sqrt{12^2+9^2}=15 yd$

Finally, the yards gained by Sam in the play are simply given by the distance covered along the forward-backward direction only. Since Sam only run from the 3 yd line to the 15 yd line along this direction, then the gain in this play was

d = 15 - 3 = 12 yd

7 0

3 years ago

How to understand that an inductor behaves like short circuit and capacitor like an open circuit in steady state in a network?

garik1379 [7]

It depends on the steady-state frequency. At zero frequency an inductor behaves like an open circuit. As the frequency increases, the inductor acts more like an open circuit and a capacitator acts more like a short circuit

8 0

3 years ago

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an

disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$ (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0

3 years ago