The field is
<em><u>E</u></em> = 1 / (4 pi ε₀) Q / <em><u>R</u></em>² directed radially outward from
the center of the shell.
R is the radius of the spherical shell.
Notice that the field is exactly the same as the field due to a point-charge
with magnitude 'Q' that's located at the center of the sphere.
The new frequency of oscillation when the car bounces on its springs is 0.447 Hz
<h3>Frequency of oscillation of spring</h3>
The frequency of oscillation of the spring is given by f = (1/2π)√(k/m) where
- k = spring constant and
- m = mass on spring
Now since k is constant, and f ∝ 1/√m.
So, we have f₂/f₁ = √(m₁/m₂) where
- f₁ = initial frequency of spring = 1.0 Hz,
- m₁ = mass of driver,
- f₂ = final frequency of spring and
- m₂ = mass on spring when driver is joined by 4 friends = 5m₁
So, making f₂ subject of the formula, we have
f₂ = [√(m₁/m₂)]f₁
Substituting the values of the variables into the equation, we have
f₂ = [√(m₁/m₂)]f₁
f₂ = [√(m₁/5m₁)]1.0 Hz
f₂ = [√(1/5)]1.0 Hz
f₂ = 1.0 Hz/√5
f₂ = 1.0 Hz/2.236
f₂ = 0.447 Hz
So, the new frequency of oscillation when the car bounces on its springs is 0.447 Hz
Learn more about frequency of oscillation of spring here:
brainly.com/question/15318845
Answer:
please read the answer below
Explanation:
The angular momentum is given by
By taking into account the angles between the vectors r and v in each case we obtain:
a)
v=(2,0)
r=(0,1)
angle = 90°
b)
r=(0,-1)
angle = 90°
c)
r=(1,0)
angle = 0°
r and v are parallel
L = 0kgm/s
d)
r=(-1,0)
angle = 180°
r and v are parallel
L = 0kgm/s
e)
r=(1,1)
angle = 45°
f)
r=(-1,1)
angle = 45°
the same as e):
L = 5kgm/s
g)
r=(-1,-1)
angle = 135°
h)
r=(1,-1)
angle = 135°
the same as g):
L = 5kgm/s
hope this helps!!
