Suppose that the cyclist begins his journey from the rest from the top of a wedge with a slope of a degree above the horizontal.
At point A (where it starts its journey), the energy is:
Ea = m * g * h
In other words, energy is only potential.
At point B (located at the bottom of the wedge), the energy is:
Eb = (1/2) * (m) * (v ^ 2)
In other words, the energy is only kinetic.
For energy conservation we have:
Ea = Eb
That is, we have that all potential energy is transformed into kinetic energy.
Which means that the cyclist has less kinetic energy at point A because that's where he has more potential energy.
answer:
the cyclist has less kinetic energy at point A because that's where he has more potential energy.
To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg.
<span>a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. </span>
<span>b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. </span>
<span>f = (ma), = .226 x 105.88, = 23.92N. </span>
Compounds are elements that are chemically combined, like water for example (it’s both hydrogen and oxygen.)
Answer:
Explanation:
reading of scale = reaction force of surface R
centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .
R = mg + m v² / R
given ,
m v² / R = .80 mg
v² = .80 x g x R
= .8 x 9.8 x 9 = 70.56
v = 8.4 m /s
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
