Answer:
CO2 increases in the winter
Explanation:
Assuming that acceleration due to gravity is g = 9.8 m/s^2.
mass of block = 5 kg
<span>μ = 0.2
</span>applied force = 9 N
The force of gravity, Fg, on this block is given by Fg = m * g --> F = 5 kg * 9.8m/s^2 = 49 N.
The normal force on this block exerted by the ground is Fn = Fg --> 49 N. (It points in direction opposite to Fg, but we don't need to worry about that here.)
The <em>maximal </em>force of friction on the block that resists its movement is Ff = Fn * <span>μ --> Ff = 49 N * 0.2 = 9.8 N
So, the max frictional force is 9.8 N.
Since the applied force is only 9 N, the block does not move.
Since we deduced that the block is not moving, then the block must be in equilibrium (all external forces must equal zero). So, the frictional force must match the applied force of 9 N.</span>
A squared plus b squared equals c squared
!ND he falls down or whatta is this a riddle or a question we can't help you if you are not clear gosh people ...
Answer:
The minimum speed the car must have at the top of the loop to not fall = 35 m/s
Explanation:
Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)
mv²/r = mg
v² = gr = 9.8 × 25 = 245
v = 15.65 m/s
But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop
Change in kinetic energy = potential energy at the top
Change in kinetic energy = (mv₂² - mv₁²)/2
v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s
v₂ = minimum velocity the car must have at the top of the loop to not fall
And potential energy at the top of the loop = mgh (where h = the diameter of the loop)
(mv₂² - mv₁²)/2 = mgh
(v₂² - v₁²) = 2gh
(v₂² - (15.65)²) = 2×9.8×50
v₂² - 245 = 980
v₂² = 1225
v₂ = 35 m/s
Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s
