Answer:

Explanation:
Acceleration on a VT graph is the slope of the line at the given point. We can find the slope at 3 with Δy/Δx. This gives us (4-2)/(3-(-3)) which works out to be -3m/s^2
Answer:
initial velocity is v = 4.95 m / s
Explanation:
To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.
Initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
y-y₀ = 0 -1/2 g t²
t = 
calculate
t =
t = 2.02 s
with this time we can substitute in the horizontal displacement equation
x = v₀ₓ t
v₀ₓ = x / t
suppose that the distance between the two points is x = 10 m
v₀ₓ = 10 / 2.02
v₀ₓ = 4.95 m / s
initial velocity is v = 4.95 m / s
Answer:
1 N
Explanation:
Buoyant Force: This is also called upthrust, It can be defined as the force which act upward exerted by a fluid when an object is placed in it.
The S.I unit is Newton.
From the question,
Buoyant force = Weight of the object in air- weight of the object when submerged in water.
U = W-W'.......................... Equation 1
Where U = upthrust, W = weight in air, W' = weight when submerged in water.
Given: W = 3 N, W' = 2 N
Substitute into equation 1
U = 3-2
W = 1 N