Answer:
Explanation:
mass of string = .0125 / 9.8
= 1.275 x 10⁻³ kg
Length of string l = 1.5 m .
m = mass per unit length
= ( .1.275 / 1.5) x 10⁻³ kg/m
m = .85 x 10⁻³ kg/m
wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
compare with equation of wave
y(x,t) = Acos(K x − ω t)
ω ( angular velocity ) = 4830 rad/s
k = 172 rad/m
Velocity = ω / k
= 4830/172 m /s
= 28.08 m /s
velocity of wave = 
28.08 = 
788.48 = W / .85 X 10⁻³
W = 670 x 10⁻³ N .
c ) wave length
wave length =2π / k
= 2 x 3.14 / 172
= .0365 m
no of wave lengths over whole length of string
= 1.5 / .0365
= 41
d )
equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by
where
m=0.1 kg is the mass of the hamburger
is the gravitational acceleration
is the increase in height of the hamburger
Substituting numbers into the equation, we find
So, the correct answer is
(3) 0.3 J
Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.
<h3>Kinetic energy</h3>
Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.
Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.
The kinetic energy is represented by the following expression:
Ec= ½ mv²
Where:
- Ec is the kinetic energy, which is measured in Joules (J).
- m is the mass measured in kilograms (kg).
- v is the speed measured in meters over seconds (m/s).
<h3>Kinetic energy of a bullet</h3>
In this case, you know:
Replacing in the definition of kinetic energy:
Ec= ½ ×0.500 kg× (25 m/s)²
Solving:
<u><em>Ec= 156.25 J</em></u>
Finally, the bullet has a kinetic energy of 156.25 J.
Learn more about kinetic energy:
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