I think its Coulomb's law<span>
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The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m
<h3>What is volume?:</h3>
This is the product of the height of a solid object and its crossectional area.
The Volume of the rod is can be calculated using the formula below.
Note: A rod has the shape of a cylinder.
Formula:
- V = πr²h............... Equation 1
Where:
- V = Volume of the rod
- r = radius of the rod
- h = height of the rod.
From the question,
Given:
- r = 4mm = 0.004 m
- h = 25 cm = 0.25 m
- π = 3.14
Substitute these values into equation 1
- V = 3.14(0.004²)(0.25)
- V = 1.26×10⁻⁵ m³
<h3>What is linear charge density:</h3>
This is the ratio of the charge on an object to the length of the object.
The linear charge density of the rod can be calculated using the formula below.
- D = Q/h.................... Equation 2
Where:
- D = Linear charge density of the rod
- Q = Charge on the rod.
- h = height or length of the rod
From the question
Given:
- Q = 0.91 C
- h = 25 cm = 0.25 m
Substitute these values into equation 2
- D = 0.91/0.25
- D = 3.64 C/m
Hence, The volume of the rod is 1.26×10⁻⁵ m³, and the linear charge density of the rod is 3.64 C/m
Learn more about charge density here: brainly.com/question/14568868
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
Answer:
U = initial velocity, t = time taken, s = distance covered. Deceleration Formula is used to calculate the deceleration of the given body in motion.
