1. A force TE 3+² +4 N is applied to the assembly starting at rest as shown. Neglect friction and all masses except for the four 3-kg particles. Determine the velocity of the particles after 5 seconds. 400 mm T= 3t² +4 3 kg T 100 mm
Answer: v(5) = 20.8 m/s
Explanation:
To solve this problem, we need to use the equation of motion for the particles in the x-direction:
F = ma
where F is the net force acting on the particles, m is the mass of the particles, and a is the acceleration of the particles.
The force acting on the particles is the tension force, TE, which is given as TE = 3t² + 4 N. Since the particles are connected by the rope, they all experience the same tension force.
The mass of each particle is 3 kg, so the total mass of the system is 12 kg.
We can now find the acceleration of the particles:
F = ma
TE = ma
a = TE/m
a = (3t² + 4 N)/(12 kg)
a = (1/4)t² + (1/3) m/s²
To find the velocity of the particles after 5 seconds, we need to integrate the acceleration with respect to time:
v = ∫a dt
v = ∫[(1/4)t² + (1/3)] dt
v = (1/12)t³ + (1/3)t + C
where C is the constant of integration. To find C, we can use the initial condition that the particles are at rest at t = 0:
v(0) = 0
C = 0
Thus, the velocity of the particles after 5 seconds is:
v(5) = (1/12)(5)³ + (1/3)(5) m/s
v(5) = 20.8 m/s
which type of marine vessel typically uses large insulated spherical tanks for product storage and has cargo piping located above the main deck?
The type of marine vessel that typically uses large insulated spherical tanks for product storage and has cargo piping located above the main deck is a liquefied gas carrier.
Liquefied gas carriers are marine vessels that are utilized for the transportation of liquefied gases such as ammonia, propane, butane, and other hazardous and non-hazardous chemicals that need low-temperature storage facilities. These carriers are constructed to transport liquefied gases under low temperature and high pressure. Liquefied gases are typically transported at a temperature of -150 degrees Celsius and a pressure of 0.3 to 1.5 MPa. Large insulated spherical tanks. The large spherical tanks used in liquefied gas carriers are a vital component of the vessel that stores the product, and they are specifically constructed to manage low-temperature storage environments. Liquefied gases are stored in large spherical tanks, which are insulated with layers of materials such as perlite, glass wool, or foam glass, to maintain a low temperature and prevent vaporization of the stored gas.
These tanks are usually located on the main deck of the ship, which is also referred to as the weather deck. Cargo piping Liquefied gas carriers have cargo piping located above the main deck, which is specifically designed to transfer the product from the storage tanks to the cargo holds or from the cargo holds to the storage tanks. The piping system is mainly made up of stainless steel, which is resistant to low temperatures and high pressures, ensuring that the product is transported securely and effectively from the storage tanks to the cargo holds.
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is the distance between the electron and the nucleus fixed for an electron in a specific orbit in the bohr model of the atom?YesNo
The distance between the electron and the nucleus is not fixed for an electron in a specific orbit in the Bohr model of the atom.
Bohr model of atomThe electron can move around the nucleus in a circular orbit, and its distance from the nucleus can vary.
The Bohr model of the atom is a simple way of visualizing the structure of the atom. In this model, the atom consists of a small, positively charged nucleus surrounded by one or more electrons in orbit. The electrons revolve around the nucleus in circular orbits of fixed energy. However, the electrons are not actually fixed in these orbits; instead, they can jump from one orbit to another by absorbing or emitting a quantum of energy.
Therefore, the distance between the electron and the nucleus is not fixed, but can change depending on the energy of the electron.
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A 4 kg box is at rest on a table. The coefficient of friction are 0.30 and 0.10 for static and kinetic respectively. Then a 10N horizontal force is applied to box.
a. What is the Normal Force acting on the box?
b. What is the value of the Friction Force?
c. What is the Net Force?
d. What is the acceleration of the box?
The Normal Force acting on the box is 39.2 N.
The Friction Force acting on the box is 11.76 N.
How to calculate the forcea. The Normal Force, denoted by N, is equal in magnitude to the weight of the box, which is given by:
N = mg
where m is the mass of the box and g is the acceleration due to gravity. Substituting the given values, we get:
N = 4 kg × 9.8 m/s² = 39.2 N
Therefore, the Normal Force acting on the box is 39.2 N.
b. The Friction Force, denoted by Ff, can be found using the formula:
Ff = μN
where μ is the coefficient of friction and N is the Normal Force. Since the box is initially at rest, the static coefficient of friction applies. Substituting the given values, we get:
Ff = 0.30 × 39.2 N = 11.76 N
Therefore, the Friction Force acting on the box is 11.76 N.
c. The Net Force, denoted by Fnet, is the sum of all the forces acting on the box. In this case, we have:
Fnet = Fapplied - Ff
where Fapplied is the applied force. Substituting the given values, we get:
Fnet = 10 N - 11.76 N = -1.76 N
The negative sign indicates that the direction of the net force is opposite to the direction of the applied force.
d. The acceleration of the box, denoted by a, can be found using the formula:
a = Fnet/m
where m is the mass of the box. Substituting the given values, we get:
a = (-1.76 N) / 4 kg = -0.44 m/s²
Therefore, the acceleration of the box is -0.44 m/s², which indicates that the box is slowing down in the direction of the applied force.
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Water flows through a horizontal bend and discharges into the atmosphere as shown. When the pressure gauge reads 10 psi, the resultant x-direction anchoring force, FAX in the horizontal plane required to hold the bend in p lace is shown in the figure. Determine the flow rate through the bend and the y-direction anchoring force, FAY required to hold the bend in place. The flow is not frictionless. Ans: 7.01 ft^3/s and 674 lbs.
The flow rate through the bend may be calculated using these equations to be 7.01 ft3/s, and the y-direction anchoring force needed to keep the bend in place is 674 pounds.
General solutions to the issue.
Step 1: Calculate the flow rate through the bend using Bernoulli's equation.
Step 2: Calculate the y-direction anchoring force necessary to hold the bend in place using the momentum equation.
Step 3: Hold the bend in place using the x-direction and y-direction anchoring forces.
The common equations that can be applied are:
Bernoulli's equation: P1 + 0.5 rhov 1 + rhogh 1 = P2 + 0.5 rhov 2 + rhogh 2 where P1 = pressure at inlet, v1 = velocity at inlet, h1 = height at inlet, P2 = pressure at outlet, v2 = velocity at outlet, h2 = height at outlet, rho = fluid density, and g = gas constant.
the speeding up caused by gravity.
FAY is the y-direction anchoring force, Q is the flow rate, v1 is the velocity at the inlet, and v2 is the velocity at the outlet. The momentum equation is FAY = rhoQ(v2 - v1).
It should be noted that the flow rate through the bend is calculated independently of the x-direction anchoring force.
The flow rate through the bend may be calculated using these equations to be 7.01 ft3/s, and the y-direction anchoring force needed to keep the bend in place is 674 pounds.
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The flow rate through the bend is 7.01 ft^3/s, and the y-direction anchoring force, FAY required to hold the bend in place is 674 lbs. in.
To solve this problem, we can use the Bernoulli equation to relate the pressure and velocity of the water at different points along the bend. We can then use the conservation of mass equation to find the flow rate through the bend.
Using the given pressure gauge reading of 10 psi, we can determine the velocity of the water at point A using the Bernoulli equation. Solving for the velocity yields 34.77 ft/s. Using this velocity and the given geometry of the bend, we can determine the area of the flow at point A as 0.2019 ft^2.
Using the conservation of mass equation, we can relate the flow rate at point A to the flow rate at point B. Solving for the flow rate at point B yields 7.01 ft^3/s.
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what value of will result in no power being dissipated by the resistor, ?
The value of the resistor that will result in no power being dissipated is when the current through the resistor is zero.
To determine this, we will use the power formula:
Power (P) = Voltage (V) x Current (I)
1: Since we want no power dissipation, set P to 0:
0 = V x I
2: To have no power dissipation, either the voltage across the resistor or the current through it must be zero. If the voltage is non-zero, then the current must be zero:
I = 0
3: According to Ohm's law, V = I x R, where R is the resistance. If I = 0, then it doesn't matter what the resistance is; the voltage will also be zero, and no power will be dissipated.
In conclusion, any value of the resistor will result in no power being dissipated if the current through the resistor is zero.
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A 4 kg box is at rest on a table. The coefficient of friction are 0.30 and 0.10 for static and kinetic respectively. Then a 10N horizontal force is applied to box.
a. What is the Normal Force acting on the box?
b. What is the value of the Friction Force?
c. What is the Net Force?
d. What is the acceleration of the box?
Answer:
Explanation:
a. normal force is mass * acceleration due to gravity, so the normal force is = 4 * 9.8, which is 39.2 N.
b. friction force is normal force * the coefficient of the friction, for static friction: .3 * 39.2 = 11.76 N. For kinetic friction: .1 * 39.2 = 3.92 N.
c. net force is the sum of all force apply on the object, since the object is rest at a table, then the only force apply on the object is the horizontal force, which is 10N, since the static force is greater than the force apply, then our net force is -1.76N( 10 - 11.76).
d. the acceleration of the box is 0 because the box is not moving.
a : normal force = 39.2N
b : static force = 11.76N, kinetic force = 3.92N
c : net force = -1.76N
d : acceleration = 0
A total of 2 metric tons of atrazine was released into the environment (20oC) after an accidental spill. The soil has the following properties: fraction of organic carbon 2.5%, bulk density 1.75 g/cm3, empirical equation, log Koc = 0.937 log Kow – 0.006. The fish has the following properties: density 2.1 g/cm3, bioconcentration factor (BCF) 8.1 cm3/g.
The volumes in the four phases are as follows:
• Volume of air: 1 x 1010 m3
• Volume of water: 8 x 108 m3
• Volume of soil: 7 x 107 m3
• Volume of fish: 4 x 104 m3
Given infinite time for the released atrazine to spread with no remedial action taken, what will be the mass and percentage of atrazine (in kg) in each of the four phases?
The percentage of atrazine in the soil phase is 1.35*10^-8
The percentage of atrazine in the soil phase is 1.45%
The percentage of atrazine in the fish phase is 6.7*10^-5
How to explain the percentageThe partition coefficient for atrazine between water and organic carbon (Koc) can be calculated using the empirical equation provided is 0.006.
The Kow value is 1.499. The distribution coefficient is 0.066.
The concentration of atrazine in water will be;
= 0.027 / 2000000 × 10^-8
= 1.35 × 20^-8
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