The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,
Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, 
: but as we saw from the previous equation, the orbital period of the Earth is proportional to , therefore as increases, T increases as well.
Therefore, the period of the orbit would increase.
Learn more about Kepler's third law:
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Answer:
Explanation:
We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for
λ
:
λ
=
v
f
Let's plug in our given values and see what we get!
λ
=
340
m
s
440
s
−
1
λ
=
0.773
m
Answer:
Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.
Maximum initial speed is 40.6 m/s
Minimum initial speed is 16.6 m/s
Explanation:
Assume this is a NET impulse so we can ignore friction.
An impulse results in a change of momentum
The impulse applied was
p = Ft = 1400(6.0) = 8400 N•s
p = mΔv
Δv = 8400 / 700 = 12 m/s
If the impulse was applied in the direction the car was already moving, the initial velocity was
vi = 28.6 - 12 = 16.6 m/s
if the impulse was applied in the direction opposite of the original velocity, the initial velocity was
vi = 28.6 + 12 = 40.6 m/s
Other angles of Net force would result in various initial velocities.
Diffuse reflection have a great day
Answer:
Energy = 7.83 x 10⁻¹⁹ J
Energy = 6.63 x 10⁻¹⁹ J
Explanation:
The energy of a photon in terms of wavelength can be calculated by the following formula:
where,
h = Plank's Constant = 6.63 x 10⁻³⁴ Js
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light
Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:
<u>Energy = 7.83 x 10⁻¹⁹ J</u>
<u></u>
Now, for λ = 300 nm = 3 x 10⁻⁷ m:
<u>Energy = 6.63 x 10⁻¹⁹ J</u>
