Answer:
Explanation:
We start converting to SI units,
A mile = 1609m
We know that the expression, which can relate linear acceleration and angular velocity is given by,
Where is the angular velocity
r=radius
linear acceleration,
Re-arrange for \omega,
Our acceleration is equal to the gravity force, so replacing,
Answer:
140.83 W
Explanation:
From the question given above, the following data were obtained:
Height (h) = 26 m
Time (t) = 12 s
Mass (m) = 6.50 Kg
Power (P) =?
Power is simply defined as the rate at which energy is used up. Mathematically, it is expressed as:
Power (P) = Energy (E) / Time (t)
But:
Energy (E) = mass (m) × acceleration due to gravity (g) × height (h)
E = mgh
P = E/t
P = mgh / t
With the above formula i.e
P = mgh / t
We can obtain the power supplied by Jill's muscles as follow:
Height (h) = 26 m
Time (t) = 12 s
Mass (m) = 6.50 Kg
Acceleration due to gravity (g) = 10 m/s²
Power (P) =?
P = mgh / t
P = 6.5 × 10 × 26 / 12
P = 140.83 W
Therefore, the power supplied by Jill's muscles is 140.83 W
Answer:
Net charge inside the cube is 321.6 C
Explanation:
Given charge density function is ρ =
side of cube is 2m
According to the definition of volume charge density total charge is given by
q = total charge = where dv = dxdydz is the volume element
on substututing the respected values we get
q =
on solving the above integration we get
q = x x
⇒ q = 321.6 C
Therefore net charge inside the cube is 321.6 C
Answer:
a) laser 1 has the maximum closest to the central maximum
b) y₂ –y₁ = L 1.66 10⁻²
Explanation:
a), B1, B2) The expression that describes the constructive interference for a double slit is
d sin θ = m λ
The pattern is observed on a screen
tan θ = y / L
Since the angles are very small
tan θ = sin θ / cos θ = sin θ = y/L
d y / L = m λ
In this case the laser has a wavelength
λ
₁ = d/20
We substitute
d y / L = m d / 20
m = 1
y₁ = L / 20
For the laser 2 λ
₂= d / 15
y₂ = L / 15
When examining the two expressions, laser 1 has the maximum closest to the central maximum
b) the difference between the two patterns is
y₂- y₁ = L (1/15 - 1/20)
y₂ –y₁ = L 1.66 10⁻²
C) laser 1 second maximum
y₁ ’= 2 L / 20
y₁ ’= L 0.1
Laser 2 third minimum
To have a minimum, the equation must be satisfied
d sin θ = (m + ½) λ
d y / L = (m + ½) λ
d y / L = (m + ½) d / 15
y = L (m +1/2) / 15
m = 3
y₂’= L (3 + ½) / 15
y₂’= L 0.2333
The difference is
y₁ ’- y₂’ = L (0.1 - 0.2333)
y₁ ’–y₂’ = L (-0.133)