Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm =
- Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
Answer:
a) 600nm
b) 300nm
Explanation:
the path difference = 2t
t = thickness of the film
L' = wavelength of light in film = L/n
L = wavength of light in air
n = refractive index of glass
(a)
for destructive interference 2t = L'/2 = L/2n
L = 4*t*n
= 4*120*10^-9*1.25
L = 600 nm
(b)
for constructive interference 2t = L' = L/1.25
L = 2tn
= 2 × 1.25 × 120nm
= 300 nm
Explanation:
It is given that,
Speed of a wave, v = 251 m/s
Wavelength of the wave, λ = 5.1 cm = 0.051 m
(1) The frequency of the wave is given by :



(2) Angular frequency of the wave is given by :



(3) The period of oscillation is given by T as :


T = 0.000203 seconds
or
T = 0.203 milliseconds
Hence, this is the required solution.
Answer:
The amount of charge the space shuttle collects is -1.224nC
Explanation:
The magnitude of Electric potential is given as;
V = kq/r
where;
V is the electric potential in volts
k is coulomb's constant
r is the radius of the sphere or distance moved by the charge
given; V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m
Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.
q = (V*r)/k
q = (-1.1 *10)/(8.99 x 10⁹ )
q = -1.224 X 10⁻⁹ C
q = -1.224nC
Therefore, the amount of charge the space shuttle collects is -1.224nC
Answer:
Diameter of the cylinder will be 
Explanation:
We have given young's modulus of steel
Change in length 
Length of rod 
Load F = 11100 KN
Strain is given by 
We know that young's modulus 
So 

We know that stress 
So 

So 