Answer:
The Forces of Flight
At any given time, there are four forces acting upon an aircraft.
These forces are lift, weight (or gravity), drag and thrust. Lift is
the key aerodynamic force that keeps objects in the air. It is the
force that opposes weight; thus, lift helps to keep an aircraft in
the air. Weight is the force that works vertically by pulling all
objects, including aircraft, toward the center of the Earth. In order
to fly an aircraft, something (lift) needs to press it in the opposite
direction of gravity. The weight of an object controls how strong
the pressure (lift) will need to be. Lift is that pressure. Drag is a
mechanical force generated by the interaction and contract of a
solid body, such as an airplane, with a fluid (liquid or gas). Finally,
the thrust is the force that is generated by the engines of an
aircraft in order for the aircraft to move forward.
Explanation:
If I were to go from the United States to China in one second, that's a large distance in an incredibly short time. I'd say that's pretty fast.
If I were to go from my room to the door of my room in a year, then that would be unbearably slow.
<u>Answer:</u> The voltage needed is 35.7 V
<u>Explanation:</u>
Assuming that the resistors are arranged in parallel combination.
For the resistors arranged in parallel combination:
We are given:
Using above equation, we get:
Calculating the voltage by using Ohm's law:
.....(1)
where,
V = voltage applied
I = Current = 3.75 A
R = Resistance = 
Putting values in equation 1, we get:
Hence, the voltage needed is 35.7 V
Answer:
v=3.66,h-3.66
Explanation:
vertical = 10sin60 - 10sin 30
horizontal =10cos60 + 10cos 30
v = 10×0.8660-10×0.5
h = 10×0.5 + 10 × 0.8660
v=8.660-5.0 = 3.66
h= 5.0-8.660 = -3.66
Answer:
<h2>
a) Q = 0.759µC</h2><h2>
b) E = 39.5µJ</h2>
Explanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
