A sand core used to form the internal surfaces of a steel casting experiences a buoyancy force of 225 n. the volume of the mold cavity forming the outside surfaces of the casting = 4840 cm3. what is the weight (in kg) of the final casting? ignore considerations of shrinkage
Newton’s second law gives us the relationship of force F, mass m and acceleration a. The formula is given as:
F = m a --> 1
However we also know that the relationship of mass m, density ρ, and volume V is:
m = V ρ --> 2
Therefore substituting equation 2 to equation 1:
F = ρ V a = ρ V g
where a is acceleration due to gravity, ρ is density of water and V is the volume of the casting, therefore:
F = (1x10^-3 kg/cm^3) (4840 cm^3) (9.8 m/s^2)
F = 47.432 kg m/s^2
F = 47.432 N
Going back to equation 1:
47.432 N = m (9.8 m/s^2)
m = 4.84 kg
Hence the weight of the final casting is 4.84 kg
To find the weight of the final casting, we use the principles of buoyancy to solve. Given the volume and the approximate density of steel, we multiply them to find the mass. The weight of the final casting is therefore approximately 37.974 kg.
Explanation:This problem's solution involves the application of Archimedes' Principle which states that an object submerged in fluid, experiences a buoyant force equal to the weight of the fluid displaced by the volume of that object.
In this circumstance, the sand core is completely submerged and the fluid (molten steel) that fills its cavity takes its form after solidifying. The given buoyancy force (225 N) is essentially the weight of the fluid displaced by it ( weight = mass x gravity). The weight of the mold cavity would hence be the weight of the steel casting.
Given the volume of the mold cavity (4840 cm
3
), we can determine the mass of the steel using the volume and the density of the steel (assuming the density of steel to be around 7.85 g/cm
3
).
The mass (m) would hence be: m = volume x density = 7.85 g/cm3 x 4840 cm3 = 37974 g = 37.974 kg.
Therefore, the weight of the final steel casting is approximately 37.974 kg.
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How the season affects the price quality and availability of vegetables?
Answer:
The season affects the price because if they aren’t grown in their natural growing season, they won’t taste as good as the other veggies would. So if they aren’t in season, and you try to buy them they will be more expensive because its harder to grow as many. So there won’t be that many, the ones that you do have will be expensive, and it’s not even going to be that good.
Explanation:
After observing the electric field in your trials above, where was the electric field the strongest? what was the direction of the electric field?
Electric field strength is highest close to central charges with closely spaced field lines, indicating intensity and weaker electric fields occur farther away.
Electric field strength: The electric field is strongest close to the central charges, where the field lines are closely spaced together. This is where the electric field is most intense. Field lines that are farther away from the central charges indicate weaker electric fields.
Convert 21.0 cm to inches
A severe storm has an average peak wave height of 16.4 feet for waves hitting the shore. suppose that a storm is in progress with a severe storm class rating. let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. if you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis?
Before we answer this question, let us first understand what alternate hypothesis is.
The alternative hypothesis is the hypothesis which is used in the hypothesis testing and this is opposite to the null hypothesis. This is the test hypothesis which is usually taken to be that the observations are the result of a real effect in an experiment.
In this case since what we want to set up is the statistical test to see if the waves are dying down, then this means we are trying to determine if the wave height are decreasing, so lesser than 16.4 feet. Therefore:
The alternative hypothesis would state (ANSWER)
Ha: μ less than 16.4 feet and P-value area is on the left of the mean.
While the null hypothesis is the opposite and would state
H0: mu equals 16.4 feet
WILL GIVE BRAINLIEST
Consider a girl that has a mass of 55.0 kg and is standing on the floor, how much supporting force does the floor have?
at least 55 kg up
at least 55 kg down
at least 539 N up
at least 539 N down
The circuit to the right is constructed using 9 identical batteries and 9 identical resistors. the lower left corner of the circuit is grounded. determine whether the current through each resistor is zero or non-zero.
Final answer:
Without seeing the specific circuit diagram, it is not possible to accurately determine whether the current through each resistor is zero or non-zero. The answer requires applying circuit analysis techniques such as Ohm's Law and the Loop Rule to the provided configuration of batteries and resistors.
Explanation:
To determine whether the current through each resistor is zero or non-zero in a circuit with 9 identical batteries and 9 identical resistors, one would need to apply principles from circuit analysis, such as Ohm's Law and the Loop Rule. However, without the provided circuit diagram, it cannot be deduced whether the current is zero or non-zero. In a typical circuit arrangement, if the circuit is properly connected with a path for the current to flow and if the batteries and resistors are arranged appropriately without any shorts, there would be a non-zero current through the resistors. The determination of this would require analyzing the specific configuration of the batteries and resistors, such as whether they are arranged in series or parallel, and applying the loop rule which states that the sum of the voltage drops in a closed loop must equal the sum of the emf sources within that loop.
how to divide electrons into energy levels?
In the figure, a 700-kg crate is on a rough surface inclined at 30°. a constant external force p = 5600 n is applied horizontally to the crate. as the force pushes the crate a distance of 3.00 m up the incline, the speed changes from 1.40 m/s to 2.50 m/s. how much work does gravity do on the crate during this process?
This problem might be confusing since there are a lot of given variable but we must take note that the problem only asks about the Work done by gravity, therefore:
Work = - Potential Energy
Where,
Potential Energy = mass * gravity * height
Note: It is negative since it is gravity doing the work on our system, not work being done on gravity.
Since the surface is inclined at 30 degrees, therefore:
height = 3 m * sin 30
height = 1.5 m
Calculating for Work:
Work = - 700 kg * 9.8 m / s^2 * 1.5 m
Work = - 10,290 J
Work = -10,300 J
Work do on the crate by gravity : W = -10290 J
Further explanationWork is the transfer of energy caused by the force acting on a moving object
Work is the product of force with the displacement of objects.
Can be formulated
W = F x SW = Work, J, Nm
F = Force, N
S = distance, m
Force and displacement of objects are vector. Multiplication of both is in accordance with the dot product concept of two vectors. Whereas W is a scalar quantity. Then if there is an angle formed by the force F and the displacement S is θ, then it can be formulated
W = F cos θ.SThere are 4 forces on the crate, so there are 4 work that occurs
work done by the applied force = Wp work done by the frictional force = Wfthe magnitude is the same as the force applied by the applied force (Fp) but the opposite direction
work done by the normal force = Wnthe direction is upward with F = mg
work done by the gravitational force = WgBecause what is sought is only work by the force of gravity, then
we can find it from the formula of potential energy
W = ΔPE = - m. g. (h2-h1)
W = - mgh2 (h1 = 0)
Negative value are caused by the movement of objects in the opposite direction to the downward force of gravity
W = - 700. 9.8. 3 sin 30 (h2 has an angle of 30°)
W = -10290 J
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In an isochoric process, the internal (thermal) energy of a gas decreases by 50 j. how much work is done by the gas during this process
In an isochoric process where the gas's internal energy decreases by 50 J, no work is done by the gas because the volume is constant. The work done is 0 J.
Explanation:In an isochoric process (also known as an isovolumetric process), the volume of a gas remains constant. According to the first law of thermodynamics, the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W): ΔU = Q - W. In an isochoric process, because the volume does not change, no work is done by the gas (W = 0). Therefore, if the internal energy of the gas decreases by 50 Joules, and no work is done, the amount of heat removed from the gas would be equivalent to the decrease in internal energy, which is -50 Joules.
The answer to the student's question is that in an isochoric process where the internal energy decreases by 50 Joules, no work is done by the gas since the volume is constant; the work done is 0 Joules.
Is it true or false that air pollution only occurs as a result of human activity
Answer:
This is false
Explanation:
The given statement "Air pollution only occurs as a result of human activity" is false because air pollution can occur as a result of both human activity and natural processes.
Human activities that contribute to air pollution include:
Burning fossil fuels: Burning fossil fuels, such as coal, oil, and natural gas, releases pollutants into the air. These pollutants include carbon dioxide, sulfur dioxide, nitrogen oxides, and particulate matter.Transportation: Transportation is another major source of air pollution. Cars, trucks, buses, and airplanes emit pollutants into the air, including carbon monoxide, hydrocarbons, and nitrogen oxides.Industry: Industrial activities, such as power plants, factories, and refineries, also emit pollutants into the air. These pollutants include sulfur dioxide, nitrogen oxides, and particulate matter.Natural processes that can contribute to air pollution include:
Volcanic eruptions: Volcanic eruptions can release large amounts of pollutants into the air, including sulfur dioxide, carbon dioxide, and particulate matter.Wildfires: Wildfires can also release large amounts of pollutants into the air, including carbon dioxide, particulate matter, and ozone.Dust storms: Dust storms can stir up dust and other particles into the air, which can contribute to air pollution.To know more about the Air pollution, here
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