Responder:
Explicación:
Usaremos la ecuación de movimiento para determinar la altura de la bola medida desde la parte superior del edificio.
Usando la ecuación para obtener la altura de caída
S = ut + 1 / 2gt²
u es la velocidad inicial = 25 m / s
g es la aceleración debida a la gravedad = 9,81 m / s²
t es el tiempo = 7 segundos
S es la altura de la caída
S = 25 (7) +1/2 (9,81) × 7²
S = 175 + 4,905 (49)
S = 175 + 240,345
S = 415,35 m
Esto significa que la pelota se elevó a 415,35 m de altura
Visible light is safe for animals because it provides them with warmth.
Hi there!
a)
We can find the angular velocity at t = 2.0 s by plugging in this value into the equation.
b)
The angular acceleration is the derivative of the angular velocity, so:
Thus, the angular acceleration is a <u>constant 25 rad/s².</u>
Use energy conservation, since no energy is lost it must be constant.
E = 0.5mv² + mgh
At release the velocity v = 0 and the height is h.
E = 0 + mgh
At impact the height h = 0 and the velocity is v.
E = 0.5mv² + 0
Since the energy E is conserved:
0.5mv² = mgh
the mass m cancels and the equation becomes:
0.5v² = gh
h = 0.5v²/g
when g = 9.81 and v = 22:
h = 24,66
Answer:
d(L)/dt = 6,96 ft/s
Explanation:
We have a right triangle where the hypotenuse (L) is the distance between the player and home plate, the legs are the line between third base and home plate ( 90 feet ) and distance between the player and third base (x) (over the line between second and third base). So we can write
L² = (90)² + x²
Applying differentiation in relation to time, on both sides of the equation we have:
2*LdL/dt = 0 + 2*x d(x)/dt (2)
In this equation we know:
d(x)/ dt = 22 feet/sec
x = 30 ft
We need to calculate L when the player is at 30 feet from third base
Then
L² = (90)² + (30)²
L² = 8100 + 900
L = √9000
L =94,87 feet
Then we are in condition for calculate d(L)/dt from the equation
2*Ld(L)/dt = 0 + 2*x d(x)/dt
2*94,87 * d(L)/dt = 2* 30* 22 ⇒ 189,74 d(L)/dt = 1320
d(L)/dt = 1320/ 189,74
d(L)/dt = 6,96 ft/s
