Answer:
θ₁ = 35.32°
Explanation:
given,
refractive index of medium 1 = n₁ = 1.75
refractive index of medium 2 = n₂ = 1.24
condition to describe the refracted angle
...(1)
Using Snell's Law
n₁ sin θ₁ = n₂ sin θ₂
θ₁ , θ₂ is the angle of incidence and refractive index
n₁. n₂ is the refractive index medium 1 and medium 2
1.75 x sin θ₁ = 1.24 x sin θ₂
From equation (1)
1.75 x sin θ₁ = 1.24 x sin (90-θ₁)
1.75 sin θ₁ = 1.24 cos θ₁
tan θ₁ = 0.708
θ₁ = 35.32°
Hence, angle of incidence is equal to θ₁ = 35.32°
Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
Answer:
114.44 J
Explanation:
From Hook's Law,
F = ke................. Equation 1
Where F = Force required to stretch the spring, k = spring constant, e = extension.
make k the subject of the equation
k = F/e.............. Equation 2
Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.
Substitute into equation 2
k = 44.5/1.016
k = 43.799 N/m
Work done in stretching the 9 in beyond its natural length
W = 1/2ke²................. Equation 3
Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m
Substitute into equation 3
W = 1/2×43.799×2.286²
W = 114.44 J
Answer:
1.4 m/s/s (2.s.f)
Explanation:
The formula for centripetal acceleration is:
, where v is velocity and r is the radius.
In the question we are given the information that the car has a mass of 1300kg, a velocity of 2.5m/s, and a turn radius of 8.5m which are all the values we need. Therefore we can simply substitute in the values to solve the question:
Therefore the centripetal acceleration of the car is 1.4m/s/s. (2.s.f)
Hope this helped!
