Answer:
Explanation:
Given
the height of the tank is
The diameter of the tank is
Density of solution
Water pressure at the bottom of the tank is
For a body falling freely from rest (disregarding air resistance), the distance the body falls varies directly as the square o
f the time. If an object is dropped from the top of a tower 490 ft high and hits the ground in 7 sec, how far did it fall in the first 5 sec?
2 answers:
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Answer:
speed of the mass is 3.546106 m / s
Explanation:
given data
mass = 77.3 g = 77.3 × kg
spring constant k = 12.5 N/m
amplitude A = 38.9 cm = 38.9 × m
to find out
the speed of the mass
solution
we will apply here conservation energy that is
K.E + P.E = Total energy ..................1
so that Total energy = K.E max = P.E max
we know amplitude so we find out first P.E max that is
PE max = K.E + P.E
(1/2)kA² = (1/2)mv² + (1/2)kx²
kA^² = mv²+ kx²
so here v² will be
v² = k(A² - x²) / m
v = √[(k/m)×(A² - x²)] ............2
here x = (1/2)A so from from 2 equation
v = √[(k/m)×(A² - (A/2)²)]
v = √[(k/m)×(3/4×A²)]
now put all value
v = √[(12.5/ 77.3 × )×(3/4×(38.9 ×
)²)]
v = 3.546106 m / s
speed of the mass is 3.546106 m / s
Answer:
The air resistance on the skydiver is 68.6 N
Explanation:
When the skydiver is falling down, there are two forces acting on him:
- The force of gravity, of magnitude , in the downward direction (where m is the mass of the skydiver and g is the acceleration due to gravity)
- The air resistance, , in the upward direction
So the net force on the skydiver is:
where
m = 7.0 kg is the mass
According to Newton's second law of motion, the net force on a body is equal to the product between its mass and its acceleration (a):
In this problem, however, the skydiver is moving with constant velocity, so his acceleration is zero:
Therefore the net force is zero:
And so, we have:
And so we can find the magnitude of the air resistance, which is equal to the force of gravity: