Complete question:
A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa
Answer:
The average pressure exerted on the surface by the block is 9655.17 Pa
Explanation:
Given;
density of the lead, ρ = 1.13 x 10⁴ kg/m³
mass of the lead block, m = 20 kg
surface area of the area of the block, A = 2.03 x 10⁻² m²
Determine the force exerted on the surface by the block due to its weight;
F = mg
F = 20 x 9.8
F = 196 N
Determine the pressure exerted on the surface by the block
P = F / A
where;
P is the pressure
P = 196 / (2.03 x 10⁻²)
P = 9655.17 N/m²
P = 9655.17 Pa
Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa
Answer:
f(x)=a(x - h)2 + k
Much like a linear function, k works like b in the slope-intercept formula. Like where add or subtract b would determine where the line crosses, in the linear, k determines the vertex of the parabola. If you're going to go up 2, then you need to add 2.
The h determines the movement horizontally. what you put in h determines if it moves left or right. To adjust this, you need to find the number to make the parentheses equal 0 when x equals -2 (because moving the vertex point to the left means subtraction/negatives):
x - h = 0
-2 - h = 0
-h = 2
h = -2
So the function ends up looking like:
f(x)=a(x - (-2))2 + 2
Subtracting a negative cancels the signs out to make a positive:
f(x)=a(x + 2)2 + 2Explanation:
Answer:
a) Ffr = -0.18 N
b) a= -1.64 m/s2
c) t = 9.2 s
d) x = 68.7 m.
e) W= -12.4 J
f) Pavg = -1.35 W
g) Pinst = -0.72 W
Explanation:
a)
- While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
- This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
- So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

where μk is the kinetic friction coefficient, and Fn is the normal force.
- Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

- We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

b)
- According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
- In this case, this net force is the friction force which we have already found in a).
- Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
- We can write the expression for a as follows:

c)
- Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

- Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

d)
- From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
- So, we can use the following kinematic equation in order to find the displacement before coming to rest:

- Since the puck comes to a stop, vf =0.
- Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

e)
- The total work done by the friction force on the object , can be obtained in several ways.
- One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
- Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

f)
- By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.

- If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

g)
- The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
- Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:
