Answer:
75degree don't forget wind and gravity force pulling down
Explanation:
Orbital speed= 2pi x radius / time period
=2pi x 1.5x10^11 / 365.25
=2.58x10^9m/day
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
Answer:4.32Nm
Explanation:
The magnitude of the torque will be the product of the force and its perpendicular distance from the force.
Force = 27N
Perpendicular distance = 16cm = 0.16m
Torque = 27×0.16
Torque = 4.32Nm
Responder:
<h2>5 horas
</h2>
Explicación:
La velocidad se define como el cambio de distancia de un cuerpo con respecto al tiempo. Matemáticamente, Velocidad = Distancia / Tiempo
Dada la velocidad del coche = 0,25 km / hy la distancia = 1,25 km
El tiempo se expresa a partir de la fórmula como Tiempo = Distancia / Velocidad
La sustitución de los valores dados en la fórmula dará;
Tiempo = 1,25 km / 0,25 km / h
Tiempo = 5 horas
Por lo tanto, la tortuga tardará 5 horas en viajar a una velocidad de 0,2 km / h.
