Answer:
The outline of the energy transfer are;
a) Kinetic energy → Clockwork spring → Potential energy
b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Please find attached the drawings of the energy transfer created with MS Visio
Explanation:
The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred
a) The energy transfer diagram for the winding up a clockwork car is given as follows;
Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;
Kinetic energy → Clockwork spring → Potential energy
b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;
Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries
Therefore, we have;
Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Answer with Explanation:
We are given that
Diameter of fighter plane=2.3 m
Radius=
a.We have to find the angular velocity in radians per second if it spins=1200 rev/min
Frequency=
1 minute=60 seconds
Angular velocity=
Angular velocity=
b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

c.Centripetal acceleration=
Centripetal acceleration==
Answer:
P = 1000000[Pa] = 1000 [kPa]
Explanation:
To solve this problem we must use the definition of pressure, which is equal to the relationship of force over area.

where:
P = pressure [Pa] (units of pascals)
F = force = 100 [N]
A = area = 100 [mm²]
But first we must convert the units from square millimeters to square meters.
![A=100[mm^{2}]*\frac{1^{2} m^{2} }{1000^{2}mm^{2} } =0.0001[m^{2} ]](https://tex.z-dn.net/?f=A%3D100%5Bmm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7D%20m%5E%7B2%7D%20%7D%7B1000%5E%7B2%7Dmm%5E%7B2%7D%20%20%7D%20%3D0.0001%5Bm%5E%7B2%7D%20%5D)
Now replacing:
![P=100/0.0001\\P=1000000[Pa]](https://tex.z-dn.net/?f=P%3D100%2F0.0001%5C%5CP%3D1000000%5BPa%5D)
Answer:
B. d(low)=4d(high)
Explanation:
Frequency of a string can be written as;
f = v/2L
Where;
v = sound velocity
L = string length
Frequency can be further expanded to;
f = v/2L = (1/2L)√(T/u) ......1
Where;
m= mass,
u = linear density of string,
T = tension
p = density of string material
A = cross sectional area of string
d = string diameter
u = m/L .......2
m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)
f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))
f = (1/2L)√(T/((p(πd^2)L/4)/L))
f = (1/2L)√(4T/pπd^2)
f = (1/L)(1/d)√(4T/pπ)
Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.
f ~ 1/d
So, if
4f(low) = f(high)
Then,
d(low) = 4d(high)
Zero maximum force (N) or field strength (N/C). ... minimum /maximum field strength.