Answer:
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Explanation:
FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE
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
Hey user
The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):
E(J) = V(V) ×<span> Q</span>(C)
So
joule = volt × coulomb
or
J = V × C
Example
What is the energy in joules that is consumed in an electrical circuit with voltage supply of 15V and charge flow of 4 coulombs?
E = 15V × 4C = 60J
Answer:
b. 14
Explanation:
= Initial temperature = 27 °C = 27 + 273 = 300 K
= Final temperature = 37 °C = 37 + 273 = 310 K
= Initial Power radiated by the object
= Final Power radiated by the object
We know that the power radiated is directly proportional to fourth power of the temperature. hence

Percentage increase in power is given as

Answer:
Earth's magnetic field serves to deflect most of the solar wind, whose charged particles would otherwise strip away the ozone layer that protects the Earth from harmful ultraviolet radiation.