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here is the question : A student throws a water balloon vertically downward from the top of a building. The balloon leaves the thrower's hand with a speed of 60.0m/s. Air resistance may be ignored,so the water balloon is in free fall after it leaves the throwers hand. a) What is its speed after falling for 2.00s? b) How far does it fall in 2.00s? c) What is the magnitude of its velocity after falling 10.0m?
Answer:
(A) 26 m/s
(B) 32.4 m
(C) v = 15.4 m/s
Explanation:
initial speed (u) = 6.4 m/s
acceleration due to gravity (a) = 9.9 m/s^[2}
time (t) = 2 s
(A) What is its speed after falling for 2.00s?
from the equation of motion v = u + at we can get the speed
v = 6.4 + (9.8 x 2) = 26 m/s
(B) How far does it fall in 2.00s?
from the equation of motion
we can get the distance covered
s = (6.4 x 2) + (0.5 x 9.8 x 2 x 2)
s = 12.8 + 19.6 = 32.4 m
c) What is the magnitude of its velocity after falling 10.0m?
from the equation of motion below we can get the velocity

v = 15.4 m/s
Wood Rots is the correct answer, as the wood begins to die
Velocity of molecules within a body
Answer:

Explanation:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

The centripetal acceleration a_c is computed as

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

The normal force N is equal to the weight of the car, thus

Equating both forces

Simplifying

Substituting the values


Keep your lane position, and sound your horn while braking in a controlled manner. Sudden panic stops are not a good idea, as they could spook the animal, causing it to suddenly dart into the path of another vehicle.