A 30 kg child on swing 2m long is released from rest when the swing supports make an angle of 37 with the vertical
a) neglecting friction, find the child's speed at lowest position
b) if the speed of the lowest position is 2.2m/s, what is the energy loss
due to friction?
We can solve this problem using conservation of energy. At the starting point A, the marble has potential energy only (mgh), and as it rolls down, some of this potential energy is converted to kinetic energy (1/2 mv^2) at any point on the curve. However, because the curve is frictionless, the total mechanical energy of the marble (potential energy plus kinetic energy) is conserved.
Let's first find the velocity of the marble at the point where it reaches a height of h/4. We can use conservation of energy between the starting point A and the point where the marble reaches a height of h/4:
mgh = 1/2 mv^2 + mgh/4
Simplifying and solving for v, we get:
v = sqrt(2gh - gh/4)
v = sqrt(7/4 * gh)
Next, we can use conservation of energy between the point where the marble reaches a height of h/4 and the highest point B:
1/2 mv^2 + mgh/4 = mgh_max
Simplifying and solving for h_max, we get:
h_max = h/4 + v^2/(2g)
h_max = h/4 + (7/8 * gh)/(2g)
h_max = h/4 + (7/16)h
h_max = 11/16 h
Therefore, the maximum height reached by the marble is 11/16 of the starting height h.
A nozzle with diameter of 0.500cm is attached to a garden hose with radius of 0.900cm. the flow rate through hose and nozzle is 0.500L/s. Calculate the speed of the water a) in the hose and b) in the nozzle
You 25kj of work on a system consisting of 3.0kg of water by stirring it with a paddle wheel. During this time, 15kcal of heat is released by the system due to poor thermal insulation. What is the change in the internal energy of the system?
a) We can use conservation of energy to find the child's speed at the lowest point. At the highest point of the swing, all of the child's potential energy (mgh) is converted into kinetic energy (1/2 mv^2) at the lowest point.
First, we need to find the height of the highest point. Since the swing is 2m long and the supports make an angle of 37 degrees with the vertical, we can use trigonometry to find the vertical component of the swing's height:
sin(37) = opposite/hypotenuse
sin(37) = h/2
h = 2sin(37)
h ≈ 1.23m
At the highest point, the child has no kinetic energy and all of their potential energy is stored as gravitational potential energy:
PE = mgh
PE = 30kg * 9.8m/s^2 * 1.23m
PE ≈ 362J
At the lowest point, all of this potential energy is converted to kinetic energy:
KE = 1/2 mv^2
362J = 1/2 * 30kg * v^2
v ≈ 4.23m/s
Therefore, the child's speed at the lowest point is approximately 4.23m/s.
b) Now we can use the given speed of 2.2m/s at the lowest point to find the energy loss due to friction. The difference between the initial and final energies is the energy loss:
initial energy = mgh = 30kg * 9.8m/s^2 * 1.23m ≈ 362J
final energy = 1/2 mv^2 = 1/2 * 30kg * (2.2m/s)^2 = 72.6J
energy loss = initial energy - final energy = 362J - 72.6J ≈ 289.4J
Therefore, the energy loss due to friction is approximately 289.4J.
A marble rolls is height h along a frictionless curved surface as shown in figure below. The projected as it reaches a height of h/4 and lands at point B. determine the maximum height reached.
We can use the equation of continuity to relate the flow rate, cross-sectional area, and speed of the water:
Q = A1v1 = A2v2
where Q is the flow rate, A is the cross-sectional area, and v is the speed of the water. Let's use subscripts 1 and 2 to denote the hose and nozzle, respectively.
a) For the hose, we can use the radius to find the cross-sectional area:
A1 = πr1^2
A1 = π(0.900cm)^2
A1 ≈ 2.54 cm^2
Using the given flow rate of 0.500 L/s, we can convert to cubic centimeters per second (cm^3/s):
Q = 0.500 L/s * 1000 cm^3/L
Q = 500 cm^3/s
Substituting into the equation of continuity, we get:
A1v1 = A2v2
v1 = A2/A1 * v2 = (πr2^2)/(πr1^2) * v2 = (r2/r1)^2 * v2
Substituting the given values, we get:
v1 = (0.500cm/0.900cm)^2 * v2
v1 ≈ 0.309v2
Substituting the flow rate and cross-sectional area into the equation of continuity, we can solve for v2:
Q = A2v2
v2 = Q/A2 = (500 cm^3/s)/(π(0.250cm)^2)
v2 ≈ 800 cm/s
Therefore, the speed of the water in the hose is:
v1 ≈ 0.309 * 800 cm/s ≈ 247 cm/s
b) For the nozzle, we can use the diameter to find the cross-sectional area:
A2 = πr2^2
A2 = π(0.250cm)^2
A2 ≈ 0.196 cm^2
Substituting into the equation of continuity, we get:
A1v1 = A2v2
v2 = A1/A2 * v1 = (πr1^2)/(πr2^2) * v1 = (r1/r2)^2 * v1
Substituting the given values, we get:
v2 = (0.900cm/0.250cm)^2 * v1
v2 ≈ 116.6v1
Substituting the flow rate and cross-sectional area into the equation of continuity, we can solve for v1:
Q = A1v1
v1 = Q/A1 = (500 cm^3/s)/(2.54 cm^2)
v1 ≈ 197 cm/s
Therefore, the speed of the water in the nozzle is:
v2 ≈ 116.6 * 197 cm/s ≈ 22,994 cm/s or 229.94 m/s (rounded to two decimal places)
A container filled with water and there are holes, as shown in the figure below. what is the ratio of x1 to x2?
Assuming the water is incompressible and neglecting the effects of viscosity, the pressure at any point in the container is the same and is equal to the atmospheric pressure. Therefore, we can use the equation of hydrostatics to relate the pressure difference between the two holes to the difference in height of the water column above them:
P_1 - P_2 = ρgh
where P is the pressure, ρ is the density of water, and h is the height of the water column above the hole. Subscript 1 denotes the hole on the left and subscript 2 denotes the hole on the right.
Since the two holes are at the same height above the bottom of the container, we can cancel out the term ρgh and get:
P_1 - P_2 = 0
which means that the pressure at the two holes is the same.
Now we can use Bernoulli's equation to relate the speed of the water at the two holes to the difference in pressure between them:
1/2 ρ v_1^2 + P_1 = 1/2 ρ v_2^2 + P_2
Since P_1 = P_2, we can simplify and get:
1/2 ρ v_1^2 = 1/2 ρ v_2^2
Canceling out the factor of 1/2 ρ, we get:
v_1^2 = v_2^2
Taking the square root of both sides, we get:
v_1 = v_2
Therefore, the velocities of water coming out of the two holes are equal, and we can conclude that x1/x2 = A2/A1, where A1 and A2 are the areas of the two holes.
From the figure, we can see that the diameters of the two holes are related as:
2x1 = x2
Therefore, the areas are related as:
A2 = π(x2/2)^2 = π(x1)^2
A1 = π(x1/2)^2 = π(x1)^2/4
The ratio of x1 to x2 is then:
x1/x2 = A2/A1 = (πx1^2)/(πx1^2/4) = 4
Therefore, the ratio of x1 to x2 is 4.
Starting with the first law of thermodynamics, we have:
ΔU = Q - W
where ΔU is the change in internal energy of the system, Q is the heat added to the system, and W is the work done on the system.
We are given that 25 kJ of work is done on the system, so W = 25 kJ.
We are also given that 15 kcal of heat is released by the system, so Q = -15 kcal. We need to convert this to joules:
1 kcal = 4.184 kJ
-15 kcal = -15 * 4.184 kJ = -62.76 kJ
Substituting these values into the first law of thermodynamics, we get:
ΔU = -62.76 kJ - 25 kJ
ΔU = -87.76 kJ
Therefore, the change in internal energy of the system is -87.76 kJ. Note that the negative sign indicates that the internal energy of the system has decreased.
During thermodynamics process, a system moves from state A to state B, it is supplied with 400J of heat and does 100J of work.
a) For this transition, what is the system's change in internal energy?
b) If the system then moves from state B back to state A, what is its
change in internal energy?
c) If in moving from A to B along different path W(AB) = 400J of work is
done on the system, How much heat does it absorbs