Answer:
17 m/s south
Explanation:
Mass of the dog = 10 kg
Mass of skateboard = 2 kg
v = Combined velocity = 2 m/s
Velocity of the dog = 1 m/s
Velocity of skateboard
In this scenario, linear momentum is conserved

The speed of the skateboard post-dog jump will be 17 m/s south, considering north as the positive direction
The required lift force is approximately 866.92 N. To determine this, we first establish the shark's mass at 92 kg and its density at 1040 kg/m³. The volume of the shark is calculated by dividing mass by density, yielding 0.08846 m³. The buoyant force acting on the shark is then determined by multiplying the volume by the density of water and gravity, resulting in a lift force of 866.92 N.
Answer:
a)n= 3.125 x
electrones.
b)J= 1.515 x
A/m²
c)
=1.114 x
m/s
d) ver explicación
Explanation:
La corriente 'I' = 5A =>5C/s
diámetro 'd'= 2.05 x
m
radio 'r' = d/2 => 1.025 x
m
número de electrones 'n'= 8.5 x
a) La cantidad de electrones que pasan por la bombilla cada segundo se determina mediante:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
Como sabemos que: Q= ne
donde e es la carga del electrón, es decir, 1.6 x
C
n= Q/e => 5/ 1.6 x 
n= 3.125 x
electrones.
b) La densidad de corriente 'J' en el cable se calcula como
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x
)²)
J= 1.515 x
A/m²
c) La velocidad típica '
' de un electrón se expresa como:
=
=1.515 x
/ 8.5 x
x |-1.6 x
|
=1.114 x
m/s
d) De acuerdo con estas ecuaciones,
J= I/A
=
=
Si utilizaras un cable de doble diámetro, ¿cuáles de las respuestas anteriores cambiarían? ¿Aumentarían o disminuirían?
V - wind speed;
53° - 35° = 18°
v² = 55² + 40² - 2 · 55 · 40 · cos 18°
v² = 3025 + 1600 - 2 · 55 · 40 · 0.951
v² = 440.6
v = √440.6
v = 20.99 ≈ 21 m/s
Conclusion: The wind speed calculates to 21 m/s.
Answer:
The outcome of adding 999mm to 100m is 101m.
Explanation:
That's my belief.