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:
Mass will be 4.437 kg
Explanation:
We have given force constant k = 7 N/m
Time period of oscillation T = 5 sec
So angular frequency 
We know that angular frequency is given by


Squaring both side

m = 4.437 kg
Explanation:
∑τ = Iα
(6 N) (0.80 m) = I (0.5 rad/s²)
I = 9.6 kg m²
Answer:
a) xf = 5.1 m
b) u = 0.304
c) x = 10.3 m
Explanation:
we will use the following formula:
u = 0.1 + A*x
Si x = 12.5 m, u = 0.6
Clearing A:
A = 0.5/12.5 = 0.04 m^-1
a) we have to:
W = Ekf - Eki
where Ekf = final kinetic energy
Eki = initial kinetic energy
9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)
Clearing xf, we have:
xf = 5.1 m
b) Replacing values for u:
u = 0.1 + (0.04*5.1) = 0.304
c) Wf = Ekf - Eki
-u*m*x*g = 0 - (m*v^2)/2
Clearing x:
x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m
Answer:
0.00001266 m
Explanation:
D = Distance from source to screen
m = Order
d = Slit separation
The distance from a point on the screen to the center line

At m = 0


At m = 1

The slit separation is 0.00001266 m