The value of x is 40°. Because, we know that a half circle equals 180 degrees. A quarter circle equals 90 degrees. 50+40=90. The reflection area has the same degree amount, so the angle is 40 °. Let me know if this helps you!
Remembering back to science I believe that temperature is on the Kelvin scale. The corresponding temperature on the Celsius scale would be very close to zero degrees which is the freezing point. The corresponding temperature on the Fahrenheit scale would be 32 degree which again is the freezing point.
Answer:
Magnitude of the Frictional force = (mv₀²)/2x₁
Explanation:
For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.
According to Newton's first law of motion
Frictional force = (mass of the box) × (deceleration experienced by the box)
Let the mass of the box be m
Then,
Frictional force = ma
Then we can obtain the deceleration using the equations of motion
v² = u² + 2ax
u = Initial velocity = v₀ m/s
v = Final velocity = 0 m/s (since the box comes to rest at the end)
x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)
a = ?
v² = u² + 2ax
0 = (v₀)² + 2ax₁
2ax₁ = - v₀²
a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)
Magnitude of the Frictional force = ma = (mv₀²)/2x₁
Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)
Answer:
μsmín = 0.1
Explanation:
- There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
- This friction force has a maximum value, that can be written as follows:

where μs is the coefficient of static friction, and Fn is the normal force,
perpendicular to the wall and aiming to the center of rotation.
- This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
- This force has the following general expression:

where ω is the angular velocity of the riders, and r the distance to the
center of rotation (the radius of the circle), and m the mass of the
riders.
Since Fc is actually Fn, we can replace the right side of (2) in (1), as
follows:

- When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

- (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
- Cancelling the masses on both sides of (4), we get:

- Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

- Replacing by the givens in (5), we can solve for μsmín, as follows:
