A) The answer is 11.53 m/s
The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi
Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²
Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h
So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h
We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m
1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s
b) The answer is 6.78 m
The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE
Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²
Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h
KE = PE
1/2 m * v² = m * a * h
Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?
1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
Answer:
3.15m³
Explanation:
To solve this problem, let us first find the mass of the petrol from the given dimension.
Mass = density x volume
Volume of petrol = 4.2m³
Density of petrol = 0.3kgm⁻³
Mass of petrol = 4.2 x 0.3 = 1.26kg
So;
We can now find the volume of the alcohol
Volume of alcohol = 
Mass of alcohol = 1.26kg
Density of alcohol = 0.4kgm⁻³
Volume of alcohol = 
= 3.15m³
Correct choices are marked in bold:
travel in straight lines and can bounce off surfaces --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces
travel through space at the speed of light --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, 
)
travel only through matter --> FALSE; electromagnetic waves can also travel through vacuum
travel only through space --> FALSE, electromagnetic waves can also travel through matter
can bend around objects --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits
move by particles bumping into each other --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved
move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave
Answer:
- quality factor (Q) = 69.99
- inductor = 1.591 x 10⁻⁴ H
- capacitor = 3.248 x 10⁻¹⁰ F
Explanation:
Given;
resonance frequency (F₀) = 700 kHz
resistor, R = 10 Ohm
bandwidth (BW) = 10 kHz
bandwidth (BW) 
make L (inductor) the subject of the formula
make C (capacitor) the subject of the formula
quality factor (Q) 
quality factor (Q) = 69.99
a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
