Answer:
because when an object approaches the speed of light, it's mass starts to increase exponentially, and would be infinite at the speed of light. It would therefore require MORE than an infinite amount of energy to accelerate even a single electron to the speed of light
Answer:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I =
)^{2}[/tex]
I =
)^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I
= 0.5 x 0.07875 x
= 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

For the object thrown in the air:
![K=\frac{1}{2}.2.[v(t)]^{2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D.2.%5Bv%28t%29%5D%5E%7B2%7D)


Kinetic energy of the object as a function of time: 
Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

For the object thrown in the air:



Potential energy as function of time: 
Total kinetic and potential energy, also known as mechanical energy is
TME =
+ (
)
TME = 1752
The expression shows that total energy of an object thrown in the air is constant and independent of time.
<span>node spacing = half of wavelength = 3 cm
velocity = 10 cm/s = freq * wavelength
hench freq = 10/6 = 5/3 = 1.7 hz</span>
Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m