Responder:
6.704 m / s
Explicación:
Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.
Trabajo realizado = Fuerza × distancia
Distancia = Trabajo realizado / Fuerza
Distancia = W / mg
S = 176/8 × 9,81
S = 176 / 78,48
S = 2,24 m
Dada la velocidad inicial u = 3.6km / h
Convertir a m / s
= 3.6km × 1000m / 1h × 3600
= 3600/3600
= 1 m / s
u = 1 m / s
Usando la ecuación de movimiento
v² = u² + 2gS para obtener la velocidad final v:
v² = 1² + 2 (9,81) (2,24)
v² = 1 + 43,9488
v² = 44,9488
v = √44,9488
v = 6,704 m / s
Por tanto, la rapidez final del perro es de 6,704 m / s
Answer:
The minimum speed when she leave the ground is 6.10 m/s.
Explanation:
Given that,
Horizontal velocity = 1.4 m/s
Height = 1.8 m
We need to calculate the minimum speed must she leave the ground
Using conservation of energy
Put the value into the formula
Hence, The minimum speed when she leave the ground is 6.10 m/s.
Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.
Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
Answer:
Explanation:
Check the attachment for solution
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
Learn more about Kinetic energy on:
brainly.com/question/12337396
#SPJ4
