Answer:
I think you just have to input the horizontal velocity component.
Explanation:
Assuming no air resistance
The horizontal velocity is constant at
vx = 17.5cos71 = 5.6974... = 5.70 i m/s
The vertical velocity varies with gravity
vy = 17.5sin71 + -9.81(22) = -199.273... = -199 j m/s
v = 5.70i - 199j m/s
arguably one should round to only two significant digits
Explanation:
the formula of speed is distance traveled by time it work
Carbon atoms and hydrogen I hope this helps
Answer:
Law of refraction
Explanation:
An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.
We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.
The vertical line is called Normal and all angles must be measured with respect to this reference in optics.
Having these angles and the refractive index of water we can use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
θ₂ =
we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.
the refractive index of air is n1 = 1 and n2= 1.33 (water)
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference

Put the value into the formula


The potential on the second plate



(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance

Put the value into the formula



Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.