The sun is bigger, but has less mass than the earth
I am pretty sure about these answers.Thermal goes in the 4th blank.
Mechanical goes in the 2nd blank.
Electrical goes in the 3rd blank.
I think chemical goes in the 1st blank and light goes in the 5th blank
Hope this helps
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
Answer:
The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°
Explanation:
According to Snell's Law,
n₁ sin θ₁ = n₂ sin θ₂
When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n
n sin θ₁ = n sin (90° - θ₁)
Note that from trigonometric relations,
Sin (90° - θ₁) = cos θ₁
n sin θ₁ = n cos θ₁
(sin θ₁)/(cos θ₁) = 1
tan θ₁ = 1
θ₁ = arctan 1 = 45°
Hope this Helps!!!
Answer:

Explanation:
<u>Horizontal Launch
</u>
It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:
- The horizontal speed is constant and equal to the initial speed

- The vertical speed is zero at launch time, but increases as the object starts to fall
- The height of the object gradually decreases until it hits the ground
- The horizontal distance where the object lands is called the range
We have the following formulas




Where
is the initial horizontal speed,
is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.
If we know the initial height of the object, we can compute the time it takes to hit the ground by using

Rearranging and solving for t



We then replace this value in

To get



The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing 

The launch velocity is
