Answer:
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
Explanation:
A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?
It decreases in speed on its way down and increases in speed on its way down.
it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center
.It increases in speed on his way down because its under the influence of gravity
from newton's equation of motion we can check by
using V^2=u^2+2as
we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level
for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.
c] Electrons
Electron gets transferred
Responder:
A. Ff = 300 N N = 784,8 N
Explicación:
Dado
Masa del cuerpo = 80 kg
Fuerza de movimiento Fm = 300N
Dado que el cuerpo no está acelerando, la fuerza de fricción (Ff) es igual a la fuerza de movimiento que actúa sobre el cuerpo, ya que la fuerza de fricción es una fuerza de oposición, es decir, Fm = Ff
Dado que Fm = 300N, Ff = 300N
La reacción normal que actúa en el cuerpo es igual al peso.
N = W = mg
g es la aceleración debida a la gravedad
g = 9,8 m / s
N = mg
N = 80 (9,81)
N = 784,8N
Por tanto, la fuerza normal que actúa sobre el cuerpo es 784,8 N
Answer:
The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is 
Explanation:
Here we have
Initial temperature of air T₁ = 20 °C = 293.15 K
Final temperature of air T₁ = 120 °C = 393.15 K
Initial pressure P₁ = 1 atm = 101325 Pa
Final pressure P₂ = Required
Area = A
Therefore we have for the pressure cooker, the volume is constant that is does not change
By Chales law
P₁/T₁ = P₂/T₂
P₂ = T₂×P₁/T₁ = 393.15 K× (101325 Pa/293.15 K) = 135,889.22 Pa
∴ P₂ = 135.88922 KPa = 135.9 kPa
Where Force = 
we have
Force = 
.
