Answer:
the density of the cube is approximately 2016.07 kg/m^3.
Explanation:
The buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
Let's first calculate the weight of the crate:
mass of crate = density * volume = density * (side length)^3 = density * 0.25^3 = 0.015625 * density
weight of crate = mass of crate * gravity = 0.015625 * density * 9.81 = 0.1530875 * density
where gravity is the acceleration due to gravity, which is approximately 9.81 m/s^2.
Since the crate is submerged in water, the buoyant force acting on it is:
buoyant force = weight of water displaced = density of water * volume of water displaced * gravity
The volume of water displaced is equal to the volume of the cube, which is 0.25^3 = 0.015625 m^3. Therefore, the buoyant force is:
buoyant force = 1000 kg/m^3 * 0.015625 m^3 * 9.81 m/s^2 = 1.534453125 N
According to the problem, it takes 310 N to lift the crate while it is still submerged. This means that the net force acting on the crate is:
net force = lifting force - buoyant force = 310 N - 1.534453125 N = 308.465546875 N
This net force is equal to the weight of the crate:
net force = weight of crate = 0.1530875 * density
Therefore, we can solve for the density of the crate:
density = net force / 0.1530875 = 308.465546875 / 0.1530875 = 2016.06666667 kg/m^3
Rounding to the nearest hundredth, we get:
density ≈ 2016.07 kg/m^3
Therefore, the density of the cube is approximately 2016.07 kg/m^3
A physical science student is building a battery with two metal plates. The table below shows a list of
metals and their electron affinity given in electron volts, eV.
Metal Electron Affinity (electron volts, eV)
Calcium - 2.87
Zinc
-0.76
Lead -0.13
Copper +0.34
The student has a copper plate. When paired with copper, which metal plate will produce a higher
voltage?
O Calcium
OZinc
O Lead
O Copper
When paired with copper, the metal plate that will produce a higher voltage is a) Calcium.
The voltage produced in a battery is related to the difference in electron affinity between the two metals used as electrodes. A greater difference in electron affinity results in a higher voltage.
In this case, the electron affinity of copper is +0.34 eV. Among the given options, calcium has the highest electron affinity with -2.87 eV. The difference between the electron affinities of copper and calcium is the greatest among the options, indicating a larger potential difference and thus a higher voltage when paired together.
Therefore, when paired with copper, the calcium plate will produce a higher voltage compared to zinc, lead, or another copper plate. Therefore, Option a is correct.
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The question was incomplete. find the full content below:
A physical science student is building a battery with two metal plates. The table below shows a list of
metals and their electron affinity given in electron volts, eV.
Metal Electron Affinity (electron volts, eV)
Calcium - 2.87
Zinc
-0.76
Lead -0.13
Copper +0.34
The student has a copper plate. When paired with copper, which metal plate will produce a higher
voltage?
A. Calcium
B. Zinc
C. Lead
D. Copper
a light string is wrapped around the rim of a small hoop if you hold the free end of the string in the hoop is released from rest it will unwind and the hoop descends, what force(s) is/are causing a torque on the hoop?
a-tension
b-weight
c-friction
d-normal force
e-more than one option is correct
Answer:
E: More than one option is correct
Explanation:
Tension, weight, and friction produce torques on the hoop, while the normal force does not.
The Sun radiates energy at a rate of about 4×1026W. At what rate is the mass decreasing?
4.44×[tex]10^{9}[/tex] kg/s is the rate at which the sun mass is decreasing.
The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.
To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.
Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.
ΔE = 4×10^26 W
c = 3×10^8 m/s (speed of light)
Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².
Substituting the values, we have:
Δm = (4×10^26 W) / (3×10^8 m/s)²
Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.
This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.
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A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 15.0°C. It is completely filled with turpentine at 15.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 85.0°C. (The average linear expansion coefficient for aluminum is 24 ✕ 10^−6°C^−1, and the average volume expansion coefficient for turpentine is 9.0 ✕ 10^−4°C−1.) Answer parts a-c.
Answer:
a) Calculate the change in the radius of the cylinder between 15.0°C and 85.0°C.
Given:
Depth of cylinder = 18.0 cm = 0.180 m
Average linear expansion coefficient for aluminum = 24 x 10^-6 /°C
Temperature change = 85 - 15 = 70 °C
Change in radius = (initial radius) x (linear expansion coefficient ) x (temperature change)
= (0.180/π) x (24 x 10^-6) x (70)
=2.16 x 10^-4 m = 0.0216 mm
b) Calculate the change in volume of the turpentine between 15.0°C and 85.0°C.
Given:
Initial volume of turpentine = 2.000 L
Average volume expansion coefficient for turpentine = 9.0 x 10^-4 /°C
Temperature change = 85 - 15 = 70 °C
Change in volume = (initial volume) x (volume expansion coefficient) x (temperature change)
= 2.000 L x (9.0 x 10^-4) x 70
= 0.126 L
c) Will any turpentine overflow? Explain your reasoning.
No turpentine will overflow because the increase in the radius of the cylinder is greater than the increase in the volume of the turpentine.
The cylinder radius increases by 0.0216 mm (part a) while the volume of turpentine increases by only 0.126 L (part b). This indicates the expanded cylinder can accommodate the increased volume of turpentine, so no overflow will occur.
Explanation:
if wrong im sorry
Joe is painting the floor of his basement using a paint roller. The roller has a mass of 2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor, Joe applies a force F = 16 N directed at an angle of 35° as shown. Ignoring the mass of the roller handle, what is the magnitude of the angular acceleration of the roller?
The magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
The magnitude of the angular acceleration of the roller can be determined using the torque equation and Newton's second law for rotational motion.
Step 1: Calculate the moment of inertia of the roller.
The moment of inertia (I) of a solid cylinder is given by the formula I = (1/2) * m * r^2, where m is the mass of the object and r is the radius.
In this case, the mass of the roller is 2.4 kg and the radius is 0.038 m.
So, I = (1/2) * 2.4 kg * (0.038 m)^2.
Step 2: Calculate the torque applied to the roller.
Torque (τ) is equal to the force (F) applied multiplied by the perpendicular distance (r) from the axis of rotation.
In this case, the force applied by Joe is 16 N and the distance is equal to the radius of the roller, 0.038 m.
So, τ = F * r.
Step 3: Use the torque equation.
The torque applied to the roller causes an angular acceleration (α) according to the equation τ = I * α.
Rearranging the equation, we get α = τ / I.
Step 4: Substitute the values into the equation.
Using the values we calculated earlier, we can substitute them into the equation α = τ / I.
α = (16 N * 0.038 m) / [(1/2) * 2.4 kg * (0.038 m)^2].
Step 5: Calculate the magnitude of the angular acceleration.
Evaluating the expression, we find that the magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
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Question 13
1.75 pts
How much heat is required to heat 0.44 kg of ice from -20 °C to water at 30°C. Numerical answer is assumed to be given in kJ.
Specific heat of ice is 2090 J/(kg "C), specific heat of water is 4186 J/(kg °C)
Latent heat of Fusion of Water is 3.33 *10^5 J/kg
Latent heat of vaporization of Water is 2.26*10^6 J/kg
The amount of heat required to heat the ice from -20 °C to water at 30°C is 238,612 J.
What is the quantity of heat required?The amount of heat required to heat the ice from -20 °C to water at 30°C is calculated as follows;
Q = Q₁ + Q₂ + Q₃
where;
Q₁ is the heat required to raise the -20⁰c to ice at 0⁰CQ₂ is the heat required to melt the ice at 0⁰CQ₃ is the heat required to raise the liquid at 0⁰C to 30⁰CThe amount of heat required to heat the ice from -20 °C to water at 30°C is calculated as;
Q = (0.44 x 4186 x 20) + (3.33 x 10⁵ x 0.44) + (0.44 x 4186 x 30)
Q = 238,612 J
Thus, the total quantity of heat required to raise the temperature of the ice to the liquid is 238,612 J.
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¿Cuál es el trabajo neto en J que se necesita para acelerar un auto de 1500 kg de 55 m/s a 65 m/s?
What is the net work in J required to accelerate a 1500 kg car from 55 m/s to 65 m/s?
The net work done (in J) required to accelerate a 1500 kg car from 55 m/s to 65 m/s is 3127500 J
How do i determine the net work done?First, we shall obtain the initial kinetic energy. Details below:
Mass (m) = 1500 Kginitial velocity (u) = 55 m/sInitial kinetic energy (KE₁) =?KE₁ = ½mu²
= ½ × 1500 × 55²
= 41250 J
Next, we shall final kinetic energy. Details below:
Mass (m) = 1500 KgFinal velocity (v) = 65 m/sFinal kinetic energy (KE₂) =?KE₂ = ½mv²
= ½ × 1500 × 65²
= 3168750 J
Finally, we shall determine the net work done. Details below:
Initial kinetic energy (KE₁) = 41250 JFinal kinetic energy (KE₂) = 3168750 JNet work done (W) =?W = KE₂ - KE₁
= 3168750 - 41250
= 3127500 J
Thus, the net work done is 3127500 J
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