What minimum speed is required for the ball to clear the 0.90-m-high net about 15.0 m from the server if the ball is "launched"
from a height of h = 2.70 m ?
1 answer:
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Answer:
= ( ρ_fluid g A) y
Explanation:
This exercise can be solved in two parts, the first finding the equilibrium force and the second finding the oscillating force
for the first part, let's write Newton's equilibrium equation
B₀ - W = 0
B₀ = W
ρ_fluid g V_fluid = W
the volume of the fluid is the area of the cube times the height it is submerged
V_fluid = A y
For the second part, the body introduces a quantity and below this equilibrium point, the equation is
B - W = m a
ρ_fluid g A (y₀ + y) - W = m a
ρ_fluid g A y + (ρ_fluid g A y₀ -W) = m a
ρ_fluid g A y + (B₀-W) = ma
the part in parentheses is zero since it is the force when it is in equilibrium
ρ_fluid g A y = m a
this equation the net force is
= ( ρ_fluid g A) y
we can see that this force varies linearly the distance and measured from the equilibrium position
Answer:
The impulse on the object is 60Ns.
Explanation:
Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.
F = m a
F = m()
⇒ Ft = m( -
)
where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, is its initialvelocity and
is the final velocity of the object.
Therefore,
impulse = Ft = m( -
)
From the question, m = 3kg, = 0m/s and
= 20m/s.
So that,
Impulse = 3 (20 - 0)
= 3(20)
= 60Ns
The impulse on the object is 60Ns.
Answer:
0.5849Weber
Explanation:
The formula for calculating the magnetic flus is expressed as:
Given
The magnitude of the magnetic field B = 3.35T
Area of the loop = πr² = 3.14(0.24)² = 0.180864m²
angle of the wire loop θ = 15.1°
Substitute the given values into the formula:
Hence the magnetic flux Φ through the loop is 0.5849Weber