Here we can use coulomb's law to find the force between two charges
As per coulombs law
]tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we have
now by using the above equation we have
so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.
Option number three is correct energy can be transformed and moved and released but it can't be destroyed and doesn't disappear.
Answer:
V (initial vertical velocity) = 45.4 sin 31.2 = 23.52 m/s
1/2 m V^2 = m g h conservation of energy
h = V^2 / (2 g) = 23.52^2 / 19.6 = 28.2 m max height
Check:
t = 28.2 / 9.8 = 2.88 sec time to reach max height
h = 23.52 * 2.88 - 1/2 g 2.88^2 = 27.1 m
Answer:
Time take to fill the standing wave to the entire length of the string is 1.3 sec.
Explanation:
Given :
The length of the one end 
, frequency of the wave 
= 2.3 Hz, wavelength of the wave λ = 1 m.
Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.
We know,
∴ 
λ
Where 
speed of the standing wave.
also, ∴ 
where 
time take to fill entire length of the string.
Compare above both equation,
⇒ 
sec
Therefore, the time taken to fill entire length 0f the string is 1.3 sec.
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
