In the experiment of free fall bob released a bag of mass 1 lb
so here we can say that initial speed of the bag is Zero
time taken by the bag to free fall is given as
t = 1.5 s
also the acceleration of free fall is given as
a = 9.8 m/s^2
now we will use kinematics equation here for finding the distance of free fall




so the bag will fall down by total distance of 11.025 m from its initial released position.
Answer:
The net acceleration of the boat is approximately 6.12 m/s² downwards
Explanation:
The buoyant or lifting force applied to the boat = 790 N
The mass of the boat lifted by the buoyant force = 214 kg
The force applied to a body is defined as the product of the mass and the acceleration of the body. Therefore, the buoyant force, F, acting on the boat can be presented as follows;
Fₐ = F - W
The weight of the boat = 214 × 9.81 = 2099.34 N
Therefore;
Fₐ = 790 - 2099.34 = -1309.34 N
Fₐ = Mass of the boat × The acceleration of the boat
Given that the buoyant force, Fₐ, is the net force acting on the boat, we have;
F = Mass of the boat × The net acceleration of the boat
F = -1309.34 N = 214 kg × The net acceleration of the boat
∴ The net acceleration of the boat = -1309.34 N/(214 kg) ≈ -6.12 m/s²
The net acceleration of the boat ≈ 6.12 m/s² downwards
<span>they had originally used einsteins therory of relativity and his field equations to figure out their information</span>
Answer:
Explanation:
Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.
Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.
Answer: 25.1 m above ground
Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s
Answer: The frog hits the ground with a velocity of 22.2 m/s