To answer this question, you must remember the equation:
a²+b²= c²
(6.4)² + (12)²= (12.2)²
<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84
This cannot be a triangle
hope this helps
Answer:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have

Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have

The block's speed is 31.422 cm/s
Answer:
Velocity
Explanation:
Velocity is an object's change in motion per unit time in a specified direction
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.
For the particular case of the Cube with equal sides its volume is determined by


In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say



In this way the net volume would be



We need to find the mass, but we have the Weight and Gravity so from Newton's second Law




PART A) From the relation of density as a unit of mass and volume we have to



PART B) To find the weight of the cube then we apply the ratio of




Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible