It is sometimes----- at night in lagos cold or colder
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m = mass of trampoline artist = 65 kg
v₀ = initial speed of the artist at the top of platform = 5
h = height through which the artist drop before landing on trampoline = 3 m
v = final speed of the artist just before landing on trampoline = ?
using conservation of energy
Kinetic energy of artist just before landing = initial kinetic energy at the top of platform + potential energy at the top of platform
(0.5) m v² = (0.5) m v₀² + mgh
dividing each term by "m"
(0.5) v² = (0.5) v₀² + gh
inserting the values
(0.5) v² = (0.5) (5)² + (9.8)(3)
v = 9.2
x = compression of the trampoline = ?
k = spring stiffness constant = 62000
Assuming the lowest depression point as reference line for measuring the potential energy
using conservation of energy
kinetic energy of artist + potential energy of artist before landing = spring potential energy of trampoline
(0.5) m v² + mg x = (0.5) k x²
inserting the values
(0.5) (65) (9.2)² + (65 x 9.8) x = (0.5) (62000) x²
x = 0.31 m
By using Ohm's law, we can calculate the resistance of the wire. Ohm's law states that:
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
Now we can use the following equation to calculate the length of the wire:
(1)
where
is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity
. The radius of the wire is half the diameter:
And the cross-sectional area is
So now we can rearrange eq.(1) to calculate the length of the wire:
where V is the potential difference across the conductor, I is the current and R the resistance. Rearranging the equation, we get
Now we can use the following equation to calculate the length of the wire:
where
L is the length of the conductor
A is its cross-sectional area
In this problem, we have a wire of copper, with resistivity
And the cross-sectional area is
So now we can rearrange eq.(1) to calculate the length of the wire:
Answer:
Due to energy loss while collision ball will not reach to same height while if there is no energy loss then in that case ball will reach to same height
Explanation:
As we know that initially ball is held at height h = 40 cm
So here we can say that kinetic energy of the ball is zero and potential energy is given as
now when strike with the ground then its its fraction of kinetic energy is lost in form of other energies
So the ball will left rebound with smaller energy and hence it will reach to height less than the initial height
While if we assume that there is no energy loss during collision then in that case ball will reach to same height again