Explanation:
At the maximum height, the ball's velocity is 0.
v² = v₀² + 2a(x - x₀)
(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)
x = 7.72 m
The ball reaches a maximum height of 7.72 m.
The times where the ball passes through half that height is:
x = x₀ + v₀ t + ½ at²
(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²
3.86 = 12.3 t - 4.9 t²
4.9 t² - 12.3 t + 3.86 = 0
Using quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8
t = 0.368, 2.14
The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.
The ship floats in water due to the buoyancy Fb that is given by the equation:
Fb=ρgV, where ρ is the density of the liquid, g=9.81 m/s² is the acceleration of the force of gravity and V is volume of the displaced liquid.
The density of fresh water is ρ₁=1000 kg/m³.
The density of salt water is in average ρ₂=1025 kg/m³.
To compare the volumes of liquids that are displaced by the ship we can take the ratio of buoyancy of salt water Fb₂ and the buoyancy of fresh water Fb₁.
The gravity force of the ship Fg=mg, where m is the mass of the ship and g=9.81 m/s², is equal to the force of buoyancy Fb₁ and Fb₂ because the mass of the ship doesn't change:
Fg=Fb₁ and Fg=Fb₂. This means Fb₁=Fb₂.
Now we can write:
Fb₂/Fb₁=(ρ₂gV₂)/(ρ₁gV₁), since Fb₁=Fb₂, they cancel out:
1/1=1=(ρ₂gV₂)/(ρ₁gV₁), g also cancels out:
(ρ₂V₂)/(ρ₁V₁)=1, now we can input ρ₁=1000 kg/m³ and ρ₂=1025 kg/m³
(1025V₂)/(1000V₁)=1
1.025(V₂/V₁)=1
V₂/V₁=1/1.025=0.9756, we multiply by V₁
V₂=0.9756V₁
Volume of salt water V₂ displaced by the ship is smaller than the volume of sweet water V₁ because the force of buoyancy of salt water is greater than the force of fresh water because salt water is more dense than fresh water.
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
Answer:
3.3m/s
Explanation:
You first get the total time (80 + 70 = 150s).
Then you would find the displacement of the truck. To do that you would do component method (vector addition), so since its a right triangle (North and East), displacement is 400^2 + 300^2 = d^2.
d= 500m.
So now that you have displacement and time, you can find the velocity:
v=d/t
v=500/150
v=3.3
Answer:a computer , machine forcery,0,push
Explanation:
