Answer:
you couldn't do this on your own or search it up on google
Answer:
c. Both forces have equal magnitudes.
Explanation:
According to Newton´s 3rd Law, the force exerted by a body on another one , is equal and opposite to the one that the second object exerts on the first one.
This does not depend on the masses of the objects, so it doesn´t matter that m₁ be greater than m₂.
As examples of this, we can mention the gravity force, the electrostatic force, the normal force, etc.
So, in this case, taking into account only the magnitudes, we can say:
F₁₂ = F₂₁
Answer: Hello, The opposite of rebel is supporter, adherent.
Explanation:
Answer:
v = 47.85 m / s
, θ = 64.7º
Explanation:
This is a missile throwing exercise.
Let's find the speed to reach the maximum height, at this point the vertical speed is zero
= v_{oy}^{2} - 2 g y
0 = v_{oy}^{2} - 2gy
v_{oy} = √2gy
let's calculate
v_{oy} = √ (2 9.8 21.3)
v_{oy} = 20.43 m / s
now we can calculate the time it takes to get to this point
vy = v_{oy} - g t
t = v_{oy} / g
t = 20.43 / 9.8
t = 2.08 s
in projectile launching, the time it takes for the body to rise is the same as the time it takes to go down, so the total launch time is
= 2 t
t_{v} = 2 2.08 = 4.16 s
let's use the horizontal throw ratio
x = v₀ₓ t_{v}
v₀ₓ = x / t_{v}
v₀ₓ = 180 / 4.16
v₀ₓ = 43.27 m / s
initial velocity is
v = √ (v₀ₓ² + v_{oy}^{2})
v = √ (20.43² + 43.27²)
v = 47.85 m / s
with an angle of
tan θ = I go / vox
θ = tan⁻¹ (43.27 / 20.43)
θ = 64.7º
Answer:
i.4.4m/s
ii.5.8m/s
III.5.98m/s
Explanation:
Given Y=70t-16t^2 at t=2
Y= 70(2)-16(2)^2=76m/s
i.at 0.1sec=2.1secs
Y=70(2.1)-16(2.1)^2=76.44m/s
Velocity (76.44-76/.1)=4.4m/s
ii.at t=0.01secs
Y=70(2.01)-16(2.01)^2=76.058m/s
Velocity=(76.056-76/.01)=5.8m/s
ii.at t=0.001
Y=70(2.001)-16(2.001)^2=76.0059m/s
Velocity=(76.0059-76/.001)=5.98m/s