We can first calculate the net force using the given information.
By Newton's second law, F(net) = ma:
F(net) = 25 * 4.3 = 107.5
We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):
f = F(net) - F(app) = 150 - 107.5 = 42.5 N
Now we can calculate the coefficient of friction, u, using the normal force, F(N):
f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
a) 120 s
b) v = 0.052R [m/s]
Explanation:
a)
The period of a revolution in a simple harmonic motion is the time taken for the object in motion to complete one cycle (in this case, the time taken to complete one revolution).
The graph of the problem is missing, find it in attachment.
To find the period of revolution of the book, we have to find the time between two consecutive points of the graph that have exactly the same shape, which correspond to two points in which the book is located at the same position.
The first point we take is t = 0, when the position of the book is x = 0.
Then, the next point with same shape is at t = 120 s, where the book returns at x = 0 m.
Therefore, the period is
T = 120 s - 0 s = 120 s
b)
The tangential speed of the book is given by the ratio between the distance covered during one revolution, which is the perimeter of the wheel, and the time taken, which is the period.
The perimeter of the wheel is:

where R is the radius of the wheel.
The period of revolution is:

Therefore, the tangential speed of the book is:

Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa
Answer:
The amount of gas that is to be released in the first second in other to attain an acceleration of 27.0 m/s2 is

Explanation:
From the question we are told that
The mass of the rocket is m = 6300 kg
The velocity at gas is being ejected is u = 2000 m/s
The initial acceleration desired is 
The time taken for the gas to be ejected is t = 1 s
Generally this desired acceleration is mathematically represented as

Here
is the rate at which gas is being ejected with respect to time
Substituting values

=> 
=> 
=> 
=> 
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m