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sergejj [24]
2 years ago
13

Which part of the wave has the highest frequency?

Physics
2 answers:
trapecia [35]2 years ago
6 0

Answer:

The last part on the right side of the diagram

Explanation:

Im on plato and just got it right :)

never [62]2 years ago
6 0

Answer:

The last part on the right side of the diagram

Explanation:

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How fast would the car need to go to double its kinetic energy?
boyakko [2]

Answer:

v_{f} = \sqrt{2}\cdot v_{o}

Explanation:

Let consider a car travelling at a speed v_{o}. The ratio of final kinetic energy to initial kinetic energy:

\frac{\frac{1}{2}\cdot m \cdot v^{2}_{f} }{\frac{1}{2}\cdot m \cdot v^{2}_{o}} = 2

\frac{v_{f}}{v_{o}} = \sqrt{2}

The final speed is:

v_{f} = \sqrt{2}\cdot v_{o}

3 0
3 years ago
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kogti [31]

Answer:

Give up.

Explanation:

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And that's the cold hard truth. :(

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1 year ago
What will happen when the sun blows up?
RSB [31]
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8 0
3 years ago
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Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac
Serjik [45]
The strength of the gravitational field is given by:
g= \frac{GM}{r^2}
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m

And now we can use the previous equation to calculate the field strength at that altitude:
g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2}  = 8.95 m/s^2

And we can see this value is a bit less than the gravitational strength at the surface, which is g_s = 9.81 m/s^2.
4 0
3 years ago
In general, fitness evaluations are meant to help you measure your physical fitness against those of similar age and gender
kirill [66]

Answer:

true

Explanation:

edge2020

8 0
3 years ago
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