3 Draw energy transfer diagrams
1 answer:
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
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The force of gravity between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the mass of the object is
, while the Earth's mass is 
. Their separation is 
, therefore the gravitational force exerted on the object is
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the mass of the object is
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t =
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
