Answer:
Explanation:
For critical angle C the relation is
SinC = 1 / μ where μ is refractive index of denser medium with respect to rarer medium .
In case of A , refractive index of denser medium with respect to rarer medium .is equal to 1.33 / 1 = 1.33
In case of B , refractive index of denser medium with respect to rarer medium .is equal to 2.42 / 1 = 2.42
In case of C , refractive index of denser medium with respect to rarer medium .is equal to 2.42 / 1.33 = 1.82
In case of D , refractive index of denser medium with respect to rarer medium .is equal to 1.50 / 1.33 = 1.127
In case of D , refractive index is lowest , critical angle will be highest .
In case of B , refractive index of denser medium with respect to rarer medium .is highest so critical angle will be lowest.
Ranking from largest to smallest
D > A >C >B .
Average speed is the speed averaged over a span of time. instantaneous speed is any speed given instant within that time
Answer:
a)= 29.4J
b)F = 588 N
c)= 60 Kg
Explanation:
Force constant of the spring (k) = 5880 N/m
Change in length of the spring (x) = 25 - 15 = 10 cm 0.1m
This work done on the spring as it is stretched (or compressed) can be recovered. This is stored work that can be used to do work on something else by this spring. That means the stretched (or compressed) spring has energy -- potential energy. This is spring potential energy or elastic potential energy.
a) Work done in pulling the body W = 1/2kx²
= 1/2 (5880)(0.1)2
= 29.4J
b)From Hook's Law,
F = ke
Where F = applied force, k = spring constant, e = extension.
Given: k = 5880 N/m, e = 25-15 = 10 cm = 0.1 m.
Substitute into the formula above
F = 5880(0.1)
F = 588 N
c)By using the formula, F = -kx
Hence mg = kx
Thus m x 9.8 =5880 x0.1
Hence mass of the body
m= 5880 x0.1/9.8
= 60 Kg
Answer:
a) D_ total = 18.54 m, b) v = 6.55 m / s
Explanation:
In this exercise we must find the displacement of the player.
a) Let's start with the initial displacement, d = 8 m at a 45º angle, use trigonometry to find the components
sin 45 = y₁ / d
cos 45 = x₁ / d
y₁ = d sin 45
x₁ = d sin 45
y₁ = 8 sin 45 = 5,657 m
x₁ = 8 cos 45 = 5,657 m
The second offset is d₂ = 12m at 90 of the 50 yard
y₂ = 12 m
x₂ = 0
total displacement
y_total = y₁ + y₂
y_total = 5,657 + 12
y_total = 17,657 m
x_total = x₁ + x₂
x_total = 5,657 + 0
x_total = 5,657 m
D_total = 17.657 i^+ 5.657 j^ m
D_total = Ra (17.657 2 + 5.657 2)
D_ total = 18.54 m
b) the average speed is requested, which is the offset carried out in the time used
v = Δx /Δt
the distance traveled using the pythagorean theorem is
r = √ (d1² + d2²)
r = √ (8² + 12²)
r = 14.42 m
The time used for this shredding is
t = t1 + t2
t = 1 + 1.2
t = 2.2 s
let's calculate the average speed
v = 14.42 / 2.2
v = 6.55 m / s