Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
1) 
The capacitance of a parallel-plate capacitor is given by:

where
is the vacuum permittivity
A is the area of each plate
d is the distance between the plates
Here, the radius of each plate is

so the area is

While the separation between the plates is

So the capacitance is

And now we can find the energy stored,which is given by:

2) 0.71 J/m^3
The magnitude of the electric field is given by

and the energy density of the electric field is given by

and using
, we find

E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
(r + h)²
where,
k = 9 × 10^9Nm²C^-2
Q = total charge, 300uC = 300 × 10^ -6C
r = 8 × 10^ -2m
h = 16 × 10^ -2m
then,
E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>
(8e^-2 + 16e^-2)²
E = 4687500N/C
Answer:
They are both correct.
Explanation:
The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.
Hence considering the both statements objectively, one can say that they are both correct.
(a) The spring stiffness constant of the spring is 18,392 N/m.
(b) The time the car was in contact with the spring before it bounces off in the opposite direction is 0.23 s.
<h3>Kinetic energy of the car</h3>
The kinetic energy of the car is calculated as follows;
K.E = ¹/₂mv²
K.E = ¹/₂ x 950 x 22²
K.E = 229,900 J
<h3>Stiffness constant of the spring</h3>
The stiffness constant of the spring is calculated as follows;
K.E = U = ¹/₂kx²
k = 2U/x²
k = (2 x 229,900)/(5)²
k = 18,392 N/m
<h3>Force exerted on the spring</h3>
F = kx
F = 18,392 x 5
F = 91,960 N
<h3>Time of impact</h3>
F = mv/t
t = mv/F
t = (950 x 22)/(91960)
t = 0.23 s
Learn more about spring constant here: brainly.com/question/1968517
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