Answer:
19.53 cm
Explanation:
The computation of the height is as follows:
Here we applied the conservation of the energy formula
As we know that
P.E of the block = P.E of the spring
m g h = ( 1 ÷ 2) k x^2
where
m = 0.15
g = 9.81
k = 420
x = 0.037
So now put the values to the above formula
(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2
1.4715 (h) = 0.28749
h = 0.19537 m
= 19.53 cm
Answer:
t = 0.1111 s
Explanation:
Let's reduce the magnitudes to the SI system
d = 120 mm (1m / 1000 mm)
d= 0.120 m
w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)
w= 56.55 rad / s
When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration
W = θ / t
t = θ / w
t = 2π / 56.55
t = 0.1111 s
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Answer:
L = 1.15 m
Explanation:
The diffraction phenomenon is described by the equation
a sin θ = m λ
Where a is the width of the slit, λ the wavelength and m is an integer, the order of diffraction is left.
The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry
tan θ = y / L
tan θ = sint θ / cos θ≈ sin θ
We substitute in the first equation
a (y / L) = m λ
The first maximum occurs for m = 1
The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is
y = 1.15 / 2 = 0.575 cm
y = 0.575 10⁻² m
Let's clear the distance to the screen (L)
L = a y / λ
Let's calculate
L = 115 10⁻⁶ 0.575 10⁻² / 575 10⁻⁹
L = 1.15 m
