Answer:
A single component that can’t be separated
brainliest please ;)
It’s either snow or rain if it’s a riddle sort of.
According to Newton's first law of motion, what happens to the ball is the ball rolls backward.
<h3>What is the first law?</h3>
This means that an object at rest or in motion will remain uniformly rectilinear and tend to be in that state if the net force on it is zero.
In this case, we have to think that the ball is at rest and the train is moving with a velocity that way, the reaction of the ball will be to go in the opposite direction to the motion.
See more about first law at brainly.com/question/3808473
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Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
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
