a) The height of the cliff is 11.68 m. b) The time taken for the rock to reach the ground when thrown straight down with the same speed is 0.8316 s (approx).
(a) Calculate the height (in m) of a cliff if it takes 2.39 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.15 m/s.
Initial velocity, u = 8.15 m/s
Final velocity, v = 0Time, t = 2.39 s
Acceleration, a = -9.8 m/s² (due to gravity)
Using the formula, s = ut + (1/2)at²
Where s is the displacement
We can get the displacement, s.
Hence, substituting the given values, we get:
s = 8.15(2.39) + (1/2)(-9.8)(2.39)²
= 11.68 m
Therefore, the height of the cliff is 11.68 m.
(b) When thrown straight down with the same speed, the initial velocity is also 8.15 m/s.
Using the formula, v = u + at
Where v is the final velocity,
u is the initial velocity,
a is the acceleration and
t is the time taken, We have:
v = 0, u = 8.15 m/s,
a = 9.8 m/s²
Hence,
0 = 8.15 + 9.8tt
= 8.15 / 9.8
= 0.8316 s
Therefore, the time taken for the rock to reach the ground when thrown straight down with the same speed is 0.8316 s (approx).
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A 25 cm x 25 cm circuit board uniformly dissipating 40 W of power is cooled by air, which
approaches the circuit board at 15°C with a velocity of 4 m/s. Disregarding any heat transfer
from the back surface of the board, determine the surface temperature of the electronic
components at the end of the board. Assume the flow to be turbulent since the electronic
components are expected to act as turbulators. Assume a film temperature of 30°C. Discuss the
validity of assumptions made to solve this problem. How does the analysis change if the film
temperature was initially assumed to be 80°C?
A higher film temperature would likely lead to a lower convective heat transfer rate and higher surface temperature for the electronic components at the end of the board.
To determine the surface temperature of the electronic components at the end of the circuit board, we can analyze the convective heat transfer between the board and the surrounding air.
Given the power dissipation (40 W), board dimensions (25 cm x 25 cm), air temperature (15°C), air velocity (4 m/s), and assuming a film temperature of 30°C, we can calculate the surface temperature.
First, we calculate the convective heat transfer coefficient (h) using empirical correlations for forced convection.
Once we have the heat transfer coefficient, we can apply Newton's law of cooling to calculate the surface temperature.
To validate the assumptions made:
Turbulent flow assumption: This assumption is reasonable since the electronic components act as turbulators, promoting turbulence in the air flow around the board.
Uniform power dissipation: Assuming uniform power dissipation across the board is common, especially if the dissipated power is evenly distributed.
If the film temperature was initially assumed to be 80°C instead of 30°C, it would affect the convective heat transfer coefficient.
Higher film temperatures usually result in lower heat transfer coefficients due to reduced temperature differences between the surface and the air.
Therefore, assuming a higher film temperature would likely lead to a lower convective heat transfer rate and higher surface temperature for the electronic components at the end of the board.
It is important to accurately estimate the film temperature to ensure accurate predictions of the surface temperature.
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0/1 pts Question 29 A hydrogen-like atom is an ion of atomic number 27 that has only one electron, What is the ion's radius in the 3rd excited state compared to the 1st Bohr radius of hydrogen atom?
The ion's radius in the 3rd excited state is 3/4 times smaller than the 1st Bohr radius of the hydrogen atom.
The ion's radius in the third excited state is calculated using the formula rn = n^2 x r1 / z, where rn is the radius of the nth orbit, r1 is the Bohr radius of hydrogen, n is the principal quantum number, and z is the atomic number.
Here, n = 3, z = 27, and r1 = 0.529 Å.
So, rn = 3^2 x 0.529 Å / 27 = 0.185 Å.
The radius of the first Bohr orbit of hydrogen is 0.529 Å.
Therefore, the ion's radius in the 3rd excited state is 0.185 Å, which is 3/4 times smaller than the first Bohr radius of the hydrogen atom.
Hence, we can conclude that the ion's radius is smaller in the 3rd excited state than in the ground state.
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Use the following data to calculate the binding energy per nucleon in MeV of the Rhodium-103 nuclide 103 Mass of Rh atom = 102.905503 u Mass of proton = 1.007276 u Mass of neutron=1.008664 u Mass of electron= 0.00054858 1 u = 931.494 MeV
The following data is given Mass of Rh atom = 102.905503 u, Mass of proton = 1.007276 u, Mass of neutron = 1.008664 u and 1 u = 931.494 MeV. The first step in calculating the binding energy per nucleon in MeV of the Rhodium-103 nuclide is to determine the number of nucleons in the Rhodium-103 nuclide.How to determine the number of nucleons in the Rhodium-103 nuclide?Rhodium-103 has 45 protons since the atomic number is 45. Since the mass number (protons + neutrons) is given as 103, the number of neutrons can be calculated as follows:
103 - 45 = 58Therefore, the number of nucleons in Rhodium-103 nuclide is 103.Binding energy is the difference between the sum of the masses of nucleons and the mass of the nucleus.The mass of nucleons is the sum of the mass of protons and neutrons.Mass of nucleons = 45 x (mass of proton) + 58 x (mass of neutron) Mass of nucleons = (45 x 1.007276) + (58 x 1.008664) = 102.47171 uThe mass defect is the difference between the mass of nucleons and the mass of the nucleus.Mass defect = mass of nucleons - mass of nucleusMass defect = 102.47171 - 102.905503 = -0.433793 uThe negative sign indicates that the mass of the nucleus is less than the mass of the nucleons. The mass defect can be converted into binding energy using Einstein's equation, E = mc².Binding energy (BE) = Mass defect × c²BE = (-0.433793 u) × (931.494 MeV/u) = -404.938 MeVThe binding energy per nucleon (BEPN) can be calculated by dividing the binding energy by the number of nucleons.Binding energy per nucleon (BEPN) = Binding energy / Number of nucleonsBEPN = (-404.938 MeV) / (103) = -3.9333 MeVTherefore, the binding energy per nucleon in MeV of the Rhodium-103 nuclide is -3.9333 MeV.
About ProtonThe proton is a subatomic particle, symbol p or p⁺, with a positive electric charge +1e elementary charge and slightly less mass than a neutron. Protons and neutrons, each with a mass of about one atomic mass unit, are collectively referred to as "nucleons". heavier than electrons. Protons are so deep in the atomic nucleus that they cannot be disturbed by particles outside the atom.
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fullt explain evryhting please and thank you:)
1. Using the example from class. find the value of the capacitance reactance being used when the power factor given is \( 0.95 \). Show: (a) Power factor angle (b) Total current (c) Current of the cap
In this question, we are given the power factor as 0.95. We need to find the capacitance reactance using the given values of power factor, power factor angle, total current, and current of the cap. We will use the following formulas to solve this problem. `
cosθ = P/S, sinθ = Q/S, tanθ = Q/P, Xc = V/Ic`.Here, θ is the power factor angle, P is the real power, Q is the reactive power, S is the apparent power, Xc is the capacitance reactance, V is the voltage, and Ic is the current of the cap. Given, Power factor = 0.95(a) Power factor angle cosθ = 0.95 => θ = cos⁻¹ (0.95) = 18.19°(b) Total currentWe know that `cosθ = P/S`. Here, S = VIcosθ. Thus, `I = S/V = P/(Vcosθ)`Substituting the values, we get `I = 2 kW / (220 V × 0.95) = 9.24 A`Therefore, the total current is 9.24 A.(c)
Current of the capWe know that `tanθ = Q/P`. Here, Q = PSinθ. Also, we know that `Xc = V/Ic`. Thus, `Ic = V/Xc`.Substituting the values, we get `Ic = 220 V / Xc`Also, `Q = Ptanθ = P × tan (18.19°) = 0.348 P`.Thus, `Ic = Q / Xcω` and `I = √(Ic² + Itotal²)`Substituting the values, we get `9.24 = √(Ic² + (0.348 × 2 kW / 220 V)²)`Solving for Ic, we get `Ic = 2.2 A`.Therefore, the current of the cap is 2.2 A.
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In terms of torques, discuss the action of a claw hammer in
pulling out nails
When it comes to the torques involved in pulling out nails using a claw hammer, the following factors are critical: the length of the hammer's
handle
, the position of the nail's head, and the angle at which the hammer is
swung
.The longer the handle of a hammer, the greater the amount of torque it can generate.
When the nail's head is as close to the surface as feasible, the torque required to remove it is
minimized
. Finally, when the hammer is swung at an angle, the torque required to remove the nail is likewise minimized.Overall, the claw hammer's
design
is intended to generate torque to make it easier to pull nails out of wood.
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An ultracentrifuge accelerates from rest to 106,000rpm in 2.10 min. What is its angular acceleration in rad/s
2
. (You do not need to enter any units.) rad/s
2
Tries 0/10 What is the tangential acceleration of a point 10.0 cm from the axis of rotation? Tries 0/10 What is the radial acceleration (in m/s
2
) of a point 10.00 cm at full rpm? (You do not need to enter any units.) m/s
2
Tries 0/10 What is the radial acceleration in multiples of g of this point at full rpm? Tries 0/10
When the engine is running at maximum speed, we may calculate the angular acceleration, tangential acceleration, radial acceleration, and radial acceleration in multiples of (g).
To find the angular acceleration of the ultracentrifuge, we can use the equation:
[tex]\[\text{{Angular acceleration}} (\alpha) = \frac{{\text{{Change in angular velocity}}}}{{\text{{Change in time}}}}\][/tex]
The change in angular velocity can be calculated by converting the given final angular velocity from rpm to rad/s and subtracting the initial angular velocity, which is 0 rad/s since it starts from rest.
The change in time is given as 2.10 min, which we need to convert to seconds.
To find the tangential acceleration of a point 10.0 cm from the axis of rotation, we can use the formula:
[tex]\[\text{{Tangential acceleration}} (a_t) = r \cdot \alpha\][/tex]
where [tex]\(r\)[/tex] is the distance from the axis of rotation. In this case, [tex]\(r = 10.0 \, \text{cm}\)[/tex] or [tex]\(0.10 \, \text{m}\)[/tex] (after converting to meters).
The radial acceleration of a point at a distance of 10.00 cm at full rpm is given by:
[tex]\[\text{{Radial acceleration}} (a_r) = r \cdot \omega^2\][/tex]
where [tex]\(\omega\)[/tex] is the angular velocity in rad/s. We can convert the given rpm value to rad/s and substitute it into the equation.
To find the radial acceleration in multiples of \(g\) at full rpm, we divide the radial acceleration by the acceleration due to gravity [tex](\(g \approx 9.8 \, \text{m/s}^2\))[/tex] and express it as a ratio.
By calculating these values using the given information, we can determine the angular acceleration, tangential acceleration, radial acceleration [tex](in m/s\(^2\))[/tex], and the radial acceleration in multiples of \(g\) at full rpm.
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