C real,inverted and smaller than the object
Answer:
Superconducting materials can transport electrons with no resistance, and hence release no heat, sound, or other energy forms. Superconductivity occurs at a specific material's critical temperature (Tc). As temperature decreases, a superconducting material's resistance gradually decreases until it reaches critical temperature. At this point resistance drops off, often to zero, as shown in the graph at right.
Explanation:
Answer:
-0.55m/s
Explanation:
Given that: For the boy
Weight = 745N
Velocity = +0.35 m/s
Mass of the boy = ?
g = 9.81m/s^2
W = mg
745 = m×9.81
m = 75.94kg
For the girl
Given that:
Weight = 477 N
g = 9.81m/s^2
m = ?
W = mg
477 = m×9.81m/s^2
m = 48.62kg
To solve for the v of the girl, the two has to add up
48.62kg×v + 75.94kg×+0.35 m/s = 0
48.62v + 26.579 = 0
48.62v = - 26.579
v = -26.579/48.62
v = -0.5466
v = -0.55m/s
Hence, the velocity of the girl is -0.55m/s.
The negative sign is as a result of the two of them moving is opposite direction.
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
C. It depends on the medium
