Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s
r = 740 m
g = 9.8 m/s²
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s
According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian
1 answer:
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Answer:
24.71 mm
Explanation:
Distance is proportional to focal length, so
d∝f
which means
Magnification of first lens
and
Similarly, magnification of second lens
and
From the above equations we get
and
which means,
and
So, we get
∴ Focal length should this camera's lens is 24.71 mm
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