Answer:
During the section CD , the speed is fastest.
Explanation:
The rate of change of distance is called speed.
Speed = distance / time
Its SI unit ism/s. It is a scalar quantity.
The slope of the distance time graph is given by the speed of the object.
Here, the speed of AB is 30/3= 10 m/s .
The speed of BC is = 0 m/s
The speed of CD is (50 - 30)/(6 - 5) = 20 m/s
So, the speed is maximum during the section CD.
Answer:
Danny hits the water with kinetic energy of 5000 J.
Explanation:
Given that,
The Weight of Danny Diver,
F = 500 N
m*g= 500 N
He steps off a diving board 10 m above the water.
h=10 m
when Danny diver hits water he generates the kinetic energy.
We need to find the kinetic energy of the water.
Let kinetic energy is K.
K = m*g*h
Where g is acceleration due to gravity.
that g= 9.8 m/s^2
now substituting the values in above equation
K= (500) * 10
K= 5000 J
Hence,
he hits the water with kinetic energy of 5000 J.
Learn more about Kinetic energy here:
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Answer:
Explanation:
Given that
g=9.8m/s²
The spring constant is
k=50N/m
The length of the bungee cord is
Lo=32m
Height of bridge which one end of the bungee is tied is 91m
A steel ball of mass 92kg is attached to the other end of the bungee.
The potential energy(Us) of the steel ball before dropped from the bridge is given as
P.E= mgh
P.E= 92×9.8×91
P.E= 82045.6 J
Us= 82045.6 J
Potential energy)(Uc) of the cord is given as
Uc= ½ke²
Where 'e' is the extension
Then the extension is final height extended by cord minus height of cord
e=hf - hi
e=hf - 32
Uc= ½×50×(hf-32)²
Uc=25(hf-32)²
Using conservation of energy,
Then,
The potential energy of free fall equals the potential energy in string
Uc=Us
25(hf-32)²=82045.6
(hf-32)² = 82045.6/25
(hf-32)²=3281.825
Take square root of both sides
√(hf-32)²=√(3281.825)
hf-32=57.29
hf=57.29+32
hf=89.29m
We neglect the negative sign of the root because the string cannot compressed
Answer:
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Explanation:
Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.
Substituting the values of the variables into the equation, we have
ΔV = V₂ - V₁.
ΔV = 175 V - 33 V.
ΔV = 142 V
The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.
So, substituting the values of the variables into the equation, we have
ΔU = eΔV
ΔU = eΔV
ΔU = -1.602 × 10⁻¹⁹ C × 142 V
ΔU = -227.484 × 10⁻¹⁹ J
ΔU = -2.27484 × 10⁻²¹ J
ΔU ≅ -2.275 × 10⁻²¹ J
So, the required equation for the electric potential energy change is
ΔU = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J
Answer:
2nd one if you ask me
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