Answer:
Due to a larger number of electrons in the heavy atoms.
Explanation:
Spectral lines are caused by the emission of light by electrons when they transit from a higher energy state(excited state) to a lower energy state.
Hence, the more electrons an atom has, the more the emission and spectral lines. The less the electrons an atom has, the less the emission and spectral lines.
Therefore, heavy nuclei (which contain more electrons) such as Iron will emit more light and so will have more spectral lines than light atoms like Hydrogen and Helium.
Answer:
more electrons in heavy atoms
Explanation:
A 0.800kg block is attached to a spring with spring constant 16.0N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 34.0cm/s . What areA)The amplitude of the subsequent oscillations?B)The block's speed at the point where x= 0.250 A?
Answer:
(a) Amplitude=0.0760 m
(b) Speed=0.337 m/s
Explanation:
(a) For amplitude
We can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
[tex]k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)mv^{2}+0=0+(1/2)kA^{2}\\ A^{2}=\frac{mv^{2}}{k} \\A=\sqrt{\frac{mv^{2}}{k}}\\ A=\sqrt{\frac{m}{k} }v\\ A=\sqrt{\frac{(0.800kg)}{16N/m} }(0.34m/s)\\A=0.0760m[/tex]
(b) For Speed
Again we can use the mentioned description of the motion and the energy conservation principle to find amplitude of oscillatory motion
[tex]k_{i}+U_{i}=K_{f}+U_{f}\\ (1/2)m(v_{i})^{2}+0=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\ (1/2)m(v_{i})^{2}=(1/2)m(v_{f} )^{2}+(1/2)k(A/2)^{2}\\(1/2)m(v_{i})^{2}-(1/2)k(A/2)^{2}=(1/2)m(v_{f} )^{2}\\(1/2)[m(v_{i})^{2}-k(A/2)^{2}]=(1/2)m(v_{f} )^{2}\\(v_{f} )^{2}=1/m[m(v_{i})^{2}-k(A/2)^{2}]\\As\\x=0.250A\\(v_{f} )^{2}=(1/0.800kg)[0.800kg(0.34m/s)^{2}-(16N/m)(0.250(0.07602m)/2)^{2}\\(v_{f} )^{2}=0.1138\\ v_{f}=\sqrt{0.1138}\\ v_{f}=0.337m/s[/tex]
Final answer:
The amplitude and speed of the block at a specified position in SHM can be determined by using conservation of energy, equating the initial kinetic energy to the maximum potential energy at the amplitude, and calculating the speed via energy values at a given displacement from equilibrium.
Explanation:
Let's break down the problem step by step:
1. Amplitude of Subsequent Oscillations (A):
- When the block is hit with the hammer, it acquires an initial velocity of [tex]\(34.0 \, \text{cm/s}\)[/tex], which we'll convert to meters per second: [tex]\(v = 34.0 \, \text{cm/s} = 0.34 \, \text{m/s}\)[/tex].
- The mechanical energy of the system (block + spring) is conserved. At the maximum extension (amplitude) of the oscillation, the kinetic energy is zero.
- Therefore, the total mechanical energy at the maximum extension is equal to the potential energy stored in the spring:
[tex]\[ E = U = \frac{1}{2} k A^2 \][/tex]
- We can express the kinetic energy at the initial point as:
[tex]\[ K = \frac{1}{2} m v^2 \][/tex]
- Since the total mechanical energy is conserved, we have:
\[ E = K + U \]
[tex]\[ \frac{1}{2} k A^2 = \frac{1}{2} m v^2 \][/tex]
- Solving for the amplitude \(A\):
[tex]\[ A = \sqrt{\frac{m v^2}{k}} \][/tex]
Substituting the given values:
[tex]\[ A = \sqrt{\frac{0.800 \, \text{kg} \cdot (0.34 \, \text{m/s})^2}{16.0 \, \text{N/m}}} \][/tex]
Calculating:
[tex]\[ A \approx 0.34 \, \text{m} \][/tex]
Therefore, the amplitude of the subsequent oscillations is approximately 0.34 meters.
2. Block's Speed at [tex]\(x = 0.250A\)[/tex]:
- At any position \(x\), the mechanical energy \(E\) of the system is given by:
[tex]\[ E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 \][/tex]
- At the maximum extension (amplitude), the kinetic energy is zero, so:
[tex]\[ E = U(x = A) = \frac{1}{2} k A^2 \][/tex]
- We can find the speed of the block at any position \(x\) using the amplitude \(A\):
[tex]\[ v = \sqrt{\frac{k}{m} (A^2 - x^2)} \][/tex]
Substituting the given value [tex]\(x = 0.250A\)[/tex]:
[tex]\[ v = \sqrt{\frac{16.0 \, \text{N/m}}{0.800 \, \text{kg}} \left(0.34^2 - (0.250 \cdot 0.34)^2\right)} \][/tex]
Calculating:
[tex]\[ v \approx 0.24 \, \text{m/s} \][/tex]
Therefore, the block's speed at the point where [tex]\(x = 0.250A\)[/tex] is approximately 0.24 meters per second
A 0.23 kg mass on a spring vibrates with amplitude 25 cm and frequency 1.7 Hz. Calculate (b) the speed at which the mass passes through equilibrium and (b) the total energy of the oscillation. (Answers: 0.82 J, 2.7 m/s).
Answer:
a) 2.67 m/s
b) 0.82 J
Explanation:
Amplitude A = 25 cm = 0.25 m
The period of the motion is the inverse of the frequency
[tex]T = \frac{1}{f} = \frac{1}{1.7} = 0.588 s[/tex]
So the angular frequency
[tex]\omega = \frac{2\pi}{T} = \frac{2\pi}{0.588} = 10.68 rad/s[/tex]
The speed at the equilibrium point is the maximum speed, at
[tex]v = \omega A = 10.68 * 0.25 = 2.67 m/s[/tex]
The spring constant can be calculated using the following
[tex]\omega^2 = \frac{k}{m} = \frac{k}{0.23}[/tex]
[tex]k = 0.23\omega^2 = 0.23*10.68^2 = 26.24 N/m[/tex]
The total energy of the oscillation is
[tex]E = kA^2 / 2 = 26.24*0.25^2 / 2 = 0.82 J[/tex]
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
Answer:
A) The expression of the electric field halfway between the plates, if the plates are in the plane Y-Z is:
[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]
B) The expression for the magnitude of the electric field E₂ just in front of the plate two ends is:
[tex]|E_2|=\displaystyle \frac{q}{2LW\varepsilon_0}[/tex]
C) The charge density is:
[tex]\sigma_2=-4.2517\cdot10^{-3}C/m^2[/tex]
Completed question:
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates
B) Input an expression for the magnitude of the electric field, E₂. Just in front of plate two END
C) If plate two has a total charge of q =-1 mC, what is its charge density, σ in C/m2?
Explanation:
A) The expression of the field can be calculated as the sum of the field produced by each plate. Each plate can be modeled as 2 parallel infinite metallic planes. Because this is a capacitor connected by both ends to a battery, the external planes have null charge (the field outside the device has to be null by definition of capacitor). This means than the charge of each plate has to be distributed in the internal faces. Because this es a metallic surface and there is no external field, we can consider a uniform charge distribution (σ=cte). Therefore in this case for each plane:
[tex]\sigma_i=\displaystyle \frac{q_i}{LW}[/tex]
The field of an infinite uniform charged plane is:
[tex]\vec{E_i}=\displaystyle \frac{\sigma_i}{2\varepsilon_0}sgn(x-x_{0i})\vec{x} =\frac{q_i}{2LW\varepsilon_0}sgn(x-x_{0i})\vec{x}[/tex]
In this case, inside the capacitor, if the plate 1 is in the left and the plate 2 is in the right, the field for 0<x<d is:
[tex]\vec{E_1}\displaystyle=\frac{q}{2LW\varepsilon_0}sgn(x})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]
[tex]\vec{E_2}\displaystyle=\frac{-q}{2LW\varepsilon_0}sgn(x-d})\vec{x}=\frac{q}{2LW\varepsilon_0}\vec{x}[/tex]
[tex]\vec{E}=\vec{E_1}+\vec{E_2}[/tex]
[tex]\vec{E}=\displaystyle \frac{q}{LW\varepsilon_0}\vec{x}[/tex]
B) we already obtain the expression of the field E₂ inside the space between the plates. Even if we are asked the expression just in front of the plate and not inside, the expression for |E₂| is still de same.
C) As seen above, we already obtain the charge density expression. Therefore we only have to replace the variables for the numerical values.
The electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.
Given:
Length, L = 0.49 m
Width, W = 0.48 m
Distance, d = 0.1 m
Here:
E = electric field
σ = surface charge density
ε₀ = permittivity of free space
We must determine the surface charge density on each plate since the plates are wired to a battery and charged so that the first plate has a charge of q.
The area of the plate is:
Area of each plate (A) = L x W = 0.49 m x 0.48 m = 0.2352 m²
The surface charge density is given by:
σ = q / A
The electric field is computed as:
E = (σ / ε₀)
Hence, the electric field, E, halfway between the plates is (σ / ε₀). The electric field is written in terms of permittivity and surface charge density.
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#Complete question is:
A capacitor is created by two metal plates. The two plates have the dimensions L = 0.49 m and W = 0.48 m. The two plates are separated by a distance, d = 0.1 m, and are parallel to each other.
A) The plates are connected to a battery and charged such that the first plate has a charge of q. Write an express of the electric field, E, halfway between the plates
An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C) is 12.0 cm from the line and moving directly toward the line at 4.10×103 m/s .
a)Calculate the proton's initial kinetic energy. Express your answer with the appropriate units.
b)How close does the proton get to the line of charge? Express your answer with the appropriate units.
Final Answer:
a) The proton's initial kinetic energy is [tex]\(8.66 \times 10^{-16}\)[/tex]J.
b) The proton gets as close as 6.00 cm to the line of charge.
Explanation:
a) In part (a), the initial kinetic energy of the proton can be calculated using the formula [tex]\(KE = \frac{1}{2}mv^2\),[/tex]where [tex]\(m\)[/tex] is the mass of the proton and [tex]\(v\)[/tex] is its velocity.
Substituting the given values, we get [tex]\(KE = \frac{1}{2}(1.67 \times 10^{-27}\, \text{kg})(4.10 \times 10^3\, \text{m/s})^2\),[/tex] resulting in [tex]\(8.66 \times 10^{-16}\) J.[/tex]
b) In part (b), the proton's closest approach can be determined using the formula for electric potential energy [tex](\(PE\))[/tex] and kinetic energy [tex](\(KE\))[/tex] when the proton is momentarily at rest.
At the closest point, all the initial kinetic energy is converted to electric potential energy, so The electric potential energy is given by [tex]\(PE = \frac{k \cdot q_1 \cdot q_2}{r}\),[/tex] where [tex]\(k\)[/tex] is Coulomb's constant, [tex]\(q_1\) and \(q_2\)[/tex] are the charges, and [tex]\(r\)[/tex] is the separation distance. Substituting the known values, [tex]\(q_1 = 1.60 \times 10^{-19}\, \text{C}\), \(q_2\)[/tex] is the charge density multiplied by the length per unit length, and [tex]\(r\)[/tex] is the distance, we can solve for [tex]\(r\),[/tex] resulting in [tex]\(6.00\, \text{cm}\).[/tex]
Final answer:
The initial kinetic energy of the proton is calculated using the formula KE = 1/2 * m * v^2, yielding 1.40x10^-20 J. The question regarding how close the proton gets to the line of charge cannot be completely answered without additional details, such as the electric field strength around the linear charge.
Explanation:
The question involves calculations relating to a proton's motion in the electric field created by a linear charge. This falls under the subject of physics and includes principles of electromagnetism and kinematics, typically taught in college-level physics courses.
a) Calculate the proton's initial kinetic energy
The kinetic energy (KE) of an object moving with velocity v is given by the equation KE = 1/2 * m * v^2, where m is the mass of the object. For a proton with mass 1.67x10^-27 kg moving at 4.10x10^3 m/s, its initial kinetic energy is:
KE = 1/2 * (1.67x10^-27 kg) * (4.10x10^3 m/s)^2 = 1.40x10^-20 J.
b) How close does the proton get to the line of charge?
This part requires the concept of energy conservation and electrostatic force. However, without specifying the potential energy due to the proton's interaction with the linear charge, the question is incomplete. Usually, one would calculate the potential energy at the closest approach and set it equal to the original kinetic energy to solve for the distance. The issue requires more information, such as the electric field strength around the line of charge, to proceed with the calculation.
The largest building in the world by volume is the boeing 747 plant in Everett, Washington. It measures approximately 632 m long, 710 yards wide, and 112 ft high.
What is the cubic volume in feet, convert your result from part a to cubic meters?
Explanation:
Given that,
The dimensions of the largest building in the world is 632 m long, 710 yards wide, and 112 ft high. It basically forms a cuboid. The volume of a cuboidal shape is given by :
Since,
1 meter = 3.28084 feet
632 m = 2073.49 feet
1 yard= 3 feet
710 yards = 2130 feet
V = lbh
[tex]V=2073.49 \ ft\times 2130\ ft\times 112\ ft[/tex]
[tex]V=494651774.4\ ft^3[/tex]
[tex]V=4.94\times 10^8\ ft^3[/tex]
Also,
[tex]V=(4.94\times 10^8\ ft^3)(\dfrac{1\ m}{3.281})^3[/tex]
[tex]V=1.39\times 10^7\ m^3[/tex]
Hence, this is the required solution.
A person walks in the following pattern: 2.9 km north, then 2.8 km west, and finally 4.4 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point
Answer:
(a) Magnitude =3.18 km
(a) Angle =28.2°
Explanation:
(a) To find magnitude of this vector recognize
[tex]R_{x}=-2.8 km\\R_{y}=-1.5km[/tex]
Use Pythagorean theorem
[tex]R=\sqrt{(R_{x})^{2}+(R_{y})^{2} }\\ R=\sqrt{(-2.8)^{2}+(-1.5)^{2} }\\ R=3.18km[/tex]
(b)To find the angle use the trigonometric property
[tex]tan\alpha =\frac{opp}{adj}\\\ tan\alpha =\frac{R_{y} }{R_{x}}\\\alpha =tan^{-1}(\frac{(-1.5)}{(-2.8)})\\\alpha =28.2^{o}[/tex]
A soda can with a volume of 345 mL is 6.5 cm in diameter and has a mass of 20g. The can is half-filled with water, and when it is placed in a tub of water it is found to float upright. What length of the can is above the water level?
Answer:
0.0473m
Explanation:
345 ml = 0.000354 m3
6.5 cm = 0.065 m
20g = 0.02 kg
Since can is half filled with water, the water volume is 0.000354 / 2 = 0.000177 m cubed
Let water density be 1000kg/m3, the mass of this half-filled water is
1000*0.000177 = 0.177 kg
The total water-can system mass is 0.177 + 0.02 = 0.197 kg
For the system to stay balanced, this mass would be equal to the mass of the water displaced by the can submerged
The volume of water displaced, or submerged can is
0.197 / 1000 = 0.000197 m cubed
Then the volume of the can that is not submerged, aka above water level is
0.000354 - 0.000197 = 0.000157 m cubed
The base area of the can is
[tex]A = \pi r^2 = \pi (d/2)^2 = \pi (0.065)^2 = 0.003318 m squared[/tex]
The length of the can that is above water is
0.000157 / 0.003318 = 0.0473 m
The half-filled soda can displaces the equivalence of its own weight in water when placed in it. Half of the soda can's total volume will always be submerged since it is only half-filled i.e., half of the can's mass is displacing water.
Explanation:The subject of this problem is the principles of buoyancy and volume. A half-filled soda can placed in water will displace its own weight of the water. The length of the can above the water level can be calculated using an understanding of volume and displacement.
First, calculate the volume of the can using the formula for the volume of a cylinder V = πr²h, where r is radius and h is height. Given the diameter of the can is 6.5 cm, the radius is 3.25 cm. The height can be calculated by rearranging the volume formula to find h. We know that the can's complete volume is 345 mL, so h (full can height) = V / (πr²).
From this, we can calculate the height of the can that is submerged in water. Since the can is half-filled, it displaces half its full weight in water. So half of the can's total volume will always be submerged. Therefore, the length of the can above the water will be half the total height of the can.
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The voltage across a conductor is increasing at a rate of 2 volts/min and the resistance is decreasing at a rate of 1 ohm/min. Use I = E/R and the Chain Rule to find the rate at which the current passing through the conductor is changing when R = 20 ohms and E = 70 volts.
Answer:
3.5 amperes
Explanation:
I = E/R
I = ?
E = 70volts
R = 20 Ohms
Therefore , I = 70/20
= 3.5 amperes
A 5.00 liter balloon of gas at 25°C is cooled to 0°C. What is the new volume (liters) of the balloon?
Answer:
4.58 L.
Explanation:
Given that
V₁ = 5 L
T₁ = 25°C = 273 + 25 = 298 K
T₂ = 0°C = 273 K
The final volume = V₂
We know that ,the ideal gas equation
If the pressure of the gas is constant ,then we can say that
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]
Now by putting the values in the above equation we get
[tex]V_2=V_1\times \dfrac{T_2}{T_1}\\V_2=5\times \dfrac{273}{298}\\V_2=4.58\ L\\[/tex]
The final volume of the balloon will be 4.58 L.
A gun is fired with angle of elevation 30°. What is the muzzle speed if the maximum height of the shell is 544 m? (Round your answer to the nearest whole number. Use g ≈ 9.8 m/s2.)
The muzzle speed of the gun, when fired at an angle of elevation of 30° and reaching a maximum height of 544 m, is approximately 329 m/s.
Explanation:The physics concept here is projectile motion. The muzzle speed of the gun can be calculated using the equation for the maximum height attained by a projectile, which is given by H = (V^2 * sin^2θ) / 2g, where V represents the muzzle speed, θ is the angle of elevation, and g is the acceleration due to gravity. Rearranging for V, and substituting the given values, we get:
V = sqrt((2 * H * g) / sin^2θ) = sqrt((2 * 544 m * 9.8 m/s^2) / sin^2 30°). Since sin 30° = 0.5, this leads to V = sqrt((2 * 544 m * 9.8 m/s^2) / (0.5)^2). The resulting muzzle speed, when calculated and rounded to the nearest whole number, is 329 m/s.
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In flow over cylinders, why does the drag coefficient suddenly drop when the flow becomes turbulent?
For a body with an aerodynamic profile to reach a low resistance coefficient, the boundary layer around the body must remain attached to its surface for as long as possible. In this way, the wake produced becomes narrow. A high shape resistance results in a wide wake. In this type of bodies, if there is turbulence, the drag coefficient increases because the pressure drag appears.
However, in the case of cylinders, it happens that the separation point of the boundary layer will move towards to the rear of the body, which will reduce the size of the wake and there will reduce the magnitude of the pressure drag.
In fluid dynamics, the sudden drop in drag coefficient when flow over a cylinder becomes turbulent is due to the shift from laminar to turbulent flow. As turbulence increases and 'energizes' the slower boundary layer of fluid on the cylinder, the overall drag is reduced and the drag coefficient decreases.
Explanation:In flow over cylinders, the sudden drop in the drag coefficient when the flow becomes turbulent can be attributed to the shift from laminar to turbulent flow itself. As the speed or Reynolds number (N'R) increases, the type of flow changes, and so does the behavior of the viscous drag exerted on the moving object.
In laminar flow, layers flow without mixing and the viscous drag is proportional to speed. As the Reynolds number enters the turbulent range, the drag begins to increase according to a different rule, becoming proportional to speed squared. This turbulent flow introduces eddies and swirls that mix fluid layers.
However, beyond a point in the turbulent flow regime, the drag coefficient starts to decrease. This is because the turbulence begins to 'energize' the generally slower, boundary layer of fluid that clings to the surface of the cylinder, which reduces the overall strength of the drag created by the flow. Moreover, these energized layers of fluid effectively 'smooth out' the obstructive effect of the cylinder, leading to a sudden decrease in the drag coefficient.
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In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their "hang time"). Treat the athlete as a particle and let ymax be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height. Ignore air resistance.
The answer details the vertical velocity needed and the horizontal distance required for a basketball player to complete a jump.
Vertical velocity: To rise 0.750 m above the floor, the athlete needs a vertical velocity of 5.43 m/s.
Horizontal distance: The athlete should start his jump 2.27 m away from the basket to reach his maximum height at the same time as he reaches the basket.
At a lab investigating fire extinguisher foams, a heavy ball is accidentally dropped into a deep vat of foam from a crane 6.10 m above the foam. After entering the foam, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the foam. The ball reaches the bottom 3.20 s after it is released. How deep is the vat?
Answer:
22.8077659955 m deep
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 6.1+0^2}\\\Rightarrow v=10.9399268736\ m/s[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{10.9399268736-0}{9.81}\\\Rightarrow t=1.11518112881\ s[/tex]
Time taken to fall through the foam
[tex]3.2-1.11518112881=2.08481887119\ s[/tex]
Distance is given by
[tex]s=vt\\\Rightarrow s=10.9399268736\times 2.08481887119\\\Rightarrow s=22.8077659955\ m[/tex]
The vat is 22.8077659955 m deep
The depth of the vat obtained is 44.076 m
Data obtained from the question Height of crane above the vat = 6.10 mTime to reach the bottom of the vat from the crane = 3.20 sDepth of vat =? Determination of the height from the crane to the bottom of the vat Time to reach the bottom of the vat from the crane (t) = 3.20 sAcceleration due to gravity (g) =? Height from crane to bottom of vat (H) =?H = ½gt²
H = ½ × 9.8 × 3.2²
H = 4.9 × 10.24
H = 50.176 m
How to determine the depth of the vatHeight from crane to bottom of vat (H) = 50.176 mHeight of crane above the vat (h) = 6.10 mDepth of vat =?Depth of vat = H – h
Depth of vat = 50.176 – 6.10
Depth of vat = 44.076 m
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Two stones resembling diamonds are suspected of being fakes. To determine if the stones might be real, the mass and volume of each are measured. Both stones have the same volume, 0.15 cm3. However, stone A has a mass of 0.52 g, and stone B has a mass of 0.42 g. If diamond has a density of 3.5 g/cm3, could either of the stones be real diamonds?
Answer:
stone A is diamond.
Explanation:
given,
Volume of the two stone = 0.15 cm³
Mass of stone A = 0.52 g
Mass of stone B = 0.42 g
Density of the diamond = 3.5 g/cm³
So, to find which stone is gold we have to calculate the density of both the stone.
We know,
[tex]density[/tex][tex]\density = \dfrac{mass}{volume}[/tex]
density of stone A
[tex]\rho_A = \dfrac{0.52}{0.15}[/tex]
[tex]\rho_A = 3.467\ g/cm^3[/tex]
density of stone B.
[tex]\rho_B = \dfrac{0.42}{0.15}[/tex]
[tex]\rho_B = 2.8\ g/cm^3[/tex]
Hence, the density of the stone A is the equal to Diamond then stone A is diamond.
Neither of the stones is a real diamond because their densities, calculated using mass and volume, do not match the density of a real diamond.
Explanation:The determination of whether a stone is a real diamond or a fake can be made by calculating the density of the stone. Density is calculated as the mass of an object divided by its volume. So, for stone A, the density is 0.52 g / 0.15 cm3 = 3.47 g/cm3, and for stone B, the density is 0.42 g / 0.15 cm3 = 2.8 g/cm3. The density of a real diamond is 3.5 g/cm3. Hence, neither stone A nor stone B is a real diamond as their densities are less than the density of a real diamond.
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A 50 kg box hangs from a rope. What is the tension in the rope if: The box is at rest? The box moves up at a steady 5.0 m/s? The box has vy=5.0 m/s and is speeding up at 5.0 m/s2? The box has vy=5.0 m/s and is slowing down at 5.0 m/s2?
Answer:
(a) [tex]T_{1}=490N[/tex]
(b) [tex]T_{2}=240N[/tex]
Explanation:
For Part (a)
Given data
The box moves up at steady 5.0 m/s
The mas of box is 50 kg
As ∑Fy=T₁ - mg=0
[tex]T_{1}=mg\\T_{1}=(50kg)(9.8m/s^{2} ) \\T_{1}=490N[/tex]
For Part(b)
Given data
[tex]v_{iy}=5m/s\\ a_{y}=-5.0m/s^{2}[/tex]
As ∑Fy=T₂ - mg=ma
[tex]T_{2}=mg+ma_{y}\\T_{2}=m(g+ a_{y})\\T_{2}=50kg(9.8-5.0) \\T_{2}=240N[/tex]
The tension in the rope varies depending on whether the box is at rest, moving at a constant velocity, or accelerating. The tension equals the weight of the box when it's at rest or moving constantly, but it will be increased or decreased by the net force caused by acceleration when the box is speeding up or slowing down.
Explanation:If the box is at rest, the tension in the rope is equal to the force of gravity. We can calculate this using the formula T = mg, where m is the mass of the box and g is the acceleration due to gravity. Therefore, T = (50 kg)(9.8 m/s²) = 490 N.
When the box moves upwards with a constant velocity, the tension in the rope also equals the weight of the box (T = mg), so the tension will stay the same at 490 N.
However, when the box is speeding up, the net force is the product of mass and acceleration. In this case, acceleration = 5.0 m/s². Using the equation Fnet = ma, we find that Fnet = (50 kg)(5 m/s²) = 250 N. The total tension now includes both the tension due to the box's weight and the additional force due to the acceleration. Therefore, T = T(g) + Fnet = 490 N + 250 N = 740 N.
Lastly, when the box is slowing down at 5.0 m/s², the net force acts in the opposite direction of the initial velocity. Using the same calculations, we find Fnet = 250 N. But this force now reduces the tension originally caused by the box's weight, so the total tension in the rope becomes T = T(g) - Fnet = 490 N - 250 N = 240 N.
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A water balloon is launched at a speed of (15.0+A) m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located (18.0+B) m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.
Answer:
The question is incomplete, below is the complete question,
"A water balloon is launched at a speed of (15.0+A) m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located (18.0+B) m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures. A=12, B=2"
Answe:
12.1m
Explanation:
Below are the data given
Speed, V=(15+A) = 15+12=27m/s
Angle of projection, ∝=36 degree
Distance from building = 18+B=18+2=20m
Since the motion describe by the object is a projectile motion, and recall that in projectile motion, motion along the horizontal path has zero acceleration and motion along the vertical path is under gravity,
the Velocity along the horizontal path is define as
[tex]V_{x}=Vcos\alpha \\V_{x}=27cos36\\V_{x}=21.8m/s[/tex]
the velocity along the Vertical path is
[tex]V_{y}=Vsin\alpha \\V_{y}=27sin36 \\V_{y}=15.87m/s[/tex]
Since the horizontal distance from the point of projection to the building is 20m, we determine the time it takes to cover this distance using the simple equation of motion
[tex]Velocity=\frac{distance }{time }\\ Time,t=27/21.8\\t=1.24secs[/tex]
The distance traveled along the vertical axis is given as
[tex]y=V_{y}t-\frac{1}{2}gt^{2}\\ t=1.24secs,\\V_{y}=15.87m/s\\g=9.81m/s[/tex]
if we substitute values, we arrive at
[tex]y=15.87*1.24-\frac{1}{2}9.81*1.24^{2}\\y=19.66-7.57\\y=12.118\\y=12.1m[/tex]
Hence the water balloon hit the building at an height of 12.1m
A uniform horizontal electric field of 1.8 × 105 N/C causes a ball that is suspended from a light string to hang at an angle of 23° from the vertical. If the mass of the ball is 5.0 grams, what is the magnitude of its charge?
Answer:
[tex]1.15669\times 10^{-7}\ C[/tex]
Explanation:
[tex]\theta[/tex] = Angle with which the electric field is hung = 23°
m = Mass of ball = 5 g
E = Electric field = [tex]1.8\times 10^5\ N/C[/tex]
T = Tension
q = Charge
We have the equations
[tex]Tcos\theta=mg[/tex]
[tex]Tsin\theta=qE[/tex]
Dividing the equations
[tex]tan\theta=\dfrac{mg}{qE}\\\Rightarrow q=\dfrac{mgtan\theta}{E}\\\Rightarrow q=\dfrac{5\times 10^{-3}\times 9.81\times tan23}{1.8\times 10^5}\\\Rightarrow q=1.15669\times 10^{-7}\ C[/tex]
The magnitude of the charge is [tex]1.15669\times 10^{-7}\ C[/tex]
A parallel-plate capacitor is connected to a battery. After it becomes charged, the capacitor is disconnected from the battery and the plate separation is increased.
What happens to the potential difference between the plates?
A) More information is needed to answer this question
B) The potential difference between the plates stays the same.
C) The potential difference between the plates decreases.
D) The potential difference between the plates increases.
Answer:
D) The potential difference between the plates increases.
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor with increased distance, will have a new capacitance C1=ϵ0kA/d1
Where d1 = nd
since d1 > d
therefore n >1
n is a factor derived as a result of the increased distance
Therefore the new capacitance becomes:
C1=ϵ0A/d1
C1= ϵ0A/nd
C1= C/n -------1
Where C1 is the capacitance with increased distance.
This implies that the charge storing capacity of the capacitor with increased plate separation decreases by a factor of (1/n) compared to that of the capacitor with original distance.
Given points
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Charge before plate separation increase same as after plate separation increase
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=C/n in equation 1. Inserting this into equation 2
CV = (CV1)/n
V1 = n(CV)/C
= n V
Since n > 1 as a result of the derived new distance, the new voltage will increase
The oscilloscope is set to measure 2 volts per division on the vertical scale. The oscilloscope display a sinusoidal voltage that vertically covers 3.6 divisions from positive to negative peak. What is the peak to peak voltage of this signal
Answer: 7.2V
Explanation:
Peak values or peak to peak voltage are calculated from RMS values, which implies VP = VRMS × √2, (assuming the source is a pure sine wave).
Since it's a sinusoidal voltage and measuring from an oscilloscope, the peak to peak voltage is gotten thus:
No of division X Volts/divisions
So, 3.6 x 2V = 7.2V
Final answer:
The peak-to-peak voltage of a sinusoidal signal covering 3.6 divisions on an oscilloscope set to 2 volts per division is 7.2 volts.
Explanation:
The question involves calculating the peak-to-peak voltage of a sinusoidal signal observed on an oscilloscope where the vertical scale is set to 2 volts per division. Given that the signal covers 3.6 divisions from positive to negative peak, we calculate the peak-to-peak voltage by multiplying the number of divisions the signal spans by the voltage per division.
To find the peak-to-peak voltage, we use the formula: Peak-to-Peak Voltage = Number of Divisions × Voltage per Division. Thus, the peak-to-peak voltage of the signal is 3.6 divisions × 2 volts/division = 7.2 volts.
A proton accelerates from rest in a uniform electric field of 680 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton.
Answer:
Acceleration, [tex]a=6.51\times 10^{10}\ m/s^2[/tex]
Explanation:
Given that,
Electric field, E = 680 N/C
Speed of the proton, v = 1.3 Mm/s
We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :
[tex]qE=ma[/tex]
a and m are the acceleration and mass of the proton.
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}[/tex]
[tex]a=6.51\times 10^{10}\ m/s^2[/tex]
So, the acceleration of the proton is [tex]a=6.51\times 10^{10}\ m/s^2[/tex]. Hence, this is the required solution.
Suppose the radius of the Earth is given to be 6378.01 km. Express the circumference of the Earth in m with 5 significant figures.
Round to 5 sig figs with trailing zeros --> 40074000
The mathematical description that fits to find the circumference of a circle (Approximation we will make to the earth considering it Uniform) is,
[tex]\phi = 2\pi r[/tex]
Here,
r = Radius
The radius of the earth is 6378.01 km or 6378010m
Replacing we have that the circumference of the Earth is
[tex]\phi = 2\pi (6378010m)[/tex]
[tex]\phi = 40074000 m[/tex]
[tex]\phi = 40074*10^3 m[/tex]
Therefore the circumference of the Earth in m with 5 significant figures is [tex]40074*10^3 m[/tex] and using only trailing zeros the answer will be [tex]40074000m[/tex]
At what point in the cardiac cycle is pressure in the ventricles the highest (around 120 mm Hg in the left ventricle)?
Answer:
Ventricular systole
Explanation:
This contraction causes an increase in pressure inside the ventricles, being the highest during the entire cardiac cycle. The ejection of blood contained in them takes place. Therefore, blood is prevented from returning to the atria by increasing pressure, which closes the bicuspid and tricuspid valves.
The highest pressure in the ventricles occurs during the ventricular systole phase of the cardiac cycle, when the ventricles are contracting to pump blood out to the body.
Explanation:The pressure in the ventricles is highest during the ventricular systole phase of the cardiac cycle. At this point, the ventricles have filled up with blood and are contracting to pump this blood out into the body. This contraction greatly increases the pressure in the ventricles, leading to a peak pressure of around 120 mm Hg in the left ventricle, depending on the individual.
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A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC moves from the point x= 0.170 m , y= 0 to the point x= 0.250 m , y= 0.250 m .
How much work W is done by the electric force on the moving point charge?
Express your answer in joules. Use k = 8.99×109 N*m^2/ C^2 for Coulomb's constant: k=1/(4*pi*epsilon0)
The work done by the electric force on the moving point charge is approximately -5.09 × 10^-5 J.
Explanation:Work done by the electric force is given by the equation W = q1 * q2 * (1/r1 - 1/r2), where q1 and q2 are the charges, r1 is the initial distance, and r2 is the final distance.
In this case, q1 = 2.30 μC, q2 = -5.00 μC, r1 = 0.170 m, and r2 = 0.250 m. Plugging these values into the equation and solving for W, we get:
W = (2.30 μC) * (-5.00 μC) * [1/√(0.170^2) - 1/√(0.250^2 + 0.250^2)]
After simplifying, the work done is approximately -5.09 × 10^-5 J.
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A rock is thrown at an angle of 60∘ to the ground. If the rock lands 25m away, what was the initial speed of the rock? (Assume air resistance is negligible. Your answer should contain the gravitational constant ????.)
Answer:
[tex]v_0 = 16.82\ m/s[/tex]
Explanation:
given,
angle at which rock is thrown = 60°
rock lands at distance,d = 25 m
initial speed of rock, = ?
In horizontal direction
distance = speed x time
d = v₀ cos 60° t
25 = v₀ cos 60° t............(1)
now,
in vertical direction
displacement in vertical direction is zero
using equation of motion
[tex]s = ut +\dfrac{1}{2}gt^2[/tex]
[tex]0 =v_0 sin 60^0 t - 4.9 t^2[/tex]
[tex]v_o sin 60^0 = 4.9 t[/tex]
[tex]t = \dfrac{v_0 sin 60^0}{4.9}[/tex]
putting the value of t in equation (1)
[tex]25 = v_0 cos 60^0\times \dfrac{v_0 sin 60^0}{4.9}[/tex]
[tex]25 =\dfrac{v_0^2cos 60^0 sin 60^0}{4.9}[/tex]v
[tex]v_0^2 = 282.90[/tex]
[tex]v_0 = 16.82\ m/s[/tex]
Hence, the initial speed of the rock is equal to 16.82 m/s
A support wire is attached to a recently transplanted tree to be sure that it stays vertical. The wire is attached to the tree at a point 1.50 m from the ground and the wire is 2.00 m long. What is the angle between the tree and the support wire?
Answer:
Explanation:
Given
Wire attached to the tree at a point [tex]h=1.5\ m[/tex] from ground
Length of wire [tex]L=2\ m[/tex]
From diagram,
Using trigonometry
[tex]\sin \theta =\frac{Perpendicular}{Hypotenuse}[/tex]
[tex]\sin \theta =\frac{1.5}{2}[/tex]
[tex]\theta =48.59[/tex]
Angle between Tree and support[tex]=90-48.59=41.41^{\circ}[/tex]
To find the angle between the tree and the support wire, we can use trigonometry. Given that the wire is 2.00 m long and attached to the tree at a point 1.50 m from the ground, the angle between the tree and the support wire is 41.1 degrees.
Explanation:To find the angle between the tree and the support wire, we can use trigonometry. The wire and the ground form a right triangle, with the wire as the hypotenuse and the vertical distance from the ground to the point of attachment as the opposite side. Using the Pythagorean theorem, we can find the length of the base of the triangle, which is the distance between the point of attachment and the tree.
Given that the wire is 2.00 m long and attached to the tree at a point 1.50 m from the ground, we can calculate the length of the base using the Pythagorean theorem: square root of (2.00^2 - 1.50^2) = 1.30 m.
Now we can use the trigonometric function tangent to find the angle between the tree and the support wire: tangent(angle) = opposite/adjacent, where the opposite side is 1.30 m and the adjacent side is 1.50 m. Solving for the angle, we get: angle = arctan(1.30/1.50) = 41.1 degrees (rounded to one decimal place).
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A cart starts at x = +6.0 m and travels towards the origin with a constant speed of 2.0 m/s. What is it the exact cart position (in m) 3.0 seconds later?
Answer:
At the origin (x' = 0 m)
Explanation:
Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.
The formula of speed is given as
S = d/t................. Equation 1
Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.
make d the subject of the equation,
d = St ................. Equation 2
Given: S = 2.0 m/s, t = 3.0 s
Substitute into equation 2
d = 2(3)
d = 6 m.
From the above, the cart covered a distance of 6 m in 3 s.
The exact position of the cart = Initial position-distance covered
x' = x-d............ Equation 3
Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.
Given: x = +6.0 m, d = 6 m.
Substitute into equation 3
x' = +6-6
x' = 0 m.
Hence the cart will be at 0 m (origin) 3 s later
The cart will be at a position of 12.0 m after 3.0 seconds.
Explanation:The cart is initially at a position of +6.0 m and is moving towards the origin with a constant speed of 2.0 m/s. We can use the formula for position to find its exact position after 3.0 seconds.
The formula for position is position = initial position + (velocity × time).
Plugging the values into the formula, we get:
position = 6.0 m + (2.0 m/s × 3.0 s) = 6.0 m + 6.0 m = 12.0 m.
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A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm. Suppose the conductor initially has a net charge of +3.8 μC instead of being neutral. What is the total charge (a) on the interior and (b) on the exterior surface?
To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.
PART A)
The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:
[tex]Q_{int}=-Q1=-1.9*10^{-6} C[/tex]. This is the total charge on the inner surface of the conducting shell.
PART B)
The positive charge (of the same value) on the external surface of the conducting shell is:
[tex]Q_{ext}=+Q_1=1.9*10^{-6} C[/tex]
The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,
[tex]Q_{ext, Total}=Q_2+Q_{ext}[/tex]
[tex]Q_{ext, Total}=1.9+3.8[/tex]
[tex]Q_{ext, Total}=5.7 \mu C[/tex]
(a) The total charge on the interior of the spherical conductor is -1.9 μC.
(b) The total exterior charge of the spherical conductor is 5.7 μC.
The given parameters;
charge at the center of the hollow sphere, q = 1.9 μC inner radius of the spherical conductor, r₁ = 3.8 cmouter radius of the spherical conductor, r₂ = 5.6 cmThe total charge on the interior is calculated as follows;
[tex]Q_{int} = - 1.9 \ \mu C[/tex]
The total exterior charge is calculated as follows;
[tex]Q_{tot . \ ext} = Q + Q_2\\\\Q_{tot . \ ext} = 1.9 \ \mu C \ + \ 3.8 \ \mu C\\\\Q_{tot . \ ext} = 5.7 \ \mu C[/tex]
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A metallic sphere has a charge of +3.1 nC. A negatively charged rod has a charge of −4.0 nC. When the rod touches the sphere, 9.2×109 electrons are transferred. What are the charges of the sphere and the rod now?
Answer:
Q'sphere=2.7*10^-9 C
Q'rod=-4.7*10^-9 C
Explanation:
given data:
charge on metallic sphere Qsphere=3.1*10^-9 C ∴1n=10^-9
charge on rod Qrod =-4*10^-9 C
no of electron n= 9.2×10^9 electrons
To find:
we are asked to find the charges Q'sphere on the sphere and Q'rod on the rod after the rod touches the sphere.
solution:
the total charge transferred when the rod touches the sphere equal to the no of electrons transferred multiplied by the charge of each electron:
Q(transferred)= nq_(e)
=(9.2×10^9)(1.6×10^-19)
=-1.312×10^-9 C
because electron are negative they move from the negatively charged rod to the positively charged rod so that new charged of the sphere is:
Q'sphere =Qsphere+Q(transferred)
=(3.1*10^-9 )-(1.312×10^-9)
=2.7*10^-9 C
similarly the new charge of the rod is:
Q'rod = Qrod-Q(transferred)
= (-6*10^-9 C)-(1.312*10^-9 C)
= -4.7*10^-9 C
∴note: there maybe error in calculation but the method is correct.
Final answer:
Upon contact, a metallic sphere and a negatively charged rod share their charges until equilibrium. The total charge of -0.9 nC is equalized, with 9.2×109 electrons changing the sphere's charge to +1.628 nC and the rod's charge to -2.528 nC.
Explanation:
When two charged objects come into contact, they share their charges until equilibrium is reached. This means each object will end up with the average charge. In the case of the metallic sphere with a charge of +3.1 nC and the negatively charged rod with a charge of -4.0 nC, the total charge before contact is (-4.0 nC) + (+3.1 nC) = -0.9 nC.
Since 9.2×109 electrons are transferred, we calculate the charge transferred using the charge of one electron, which is approximately -1.6×10-19 C. Multiplying the number of electrons by the charge of one electron gives us the total charge transferred: 9.2×109 × -1.6×10-19 C/electron ≈ -1.472 nC.
This charge is added to the metallic sphere and subtracted from the rod. So, the new charge on the sphere is +3.1 nC + (-1.472 nC) = +1.628 nC, and the charge on the rod is -4.0 nC - (-1.472 nC) = -2.528 nC. Both charges are now closer in magnitude, representing the sharing of charges due to contact.
A 220 g , 23-cm-diameter plastic disk is spun on an axle through its center by an electric motor.What torque must the motor supply to take the disk from 0 to 1800 rpm in 4.6 s ?
The torque required by the motor to spin a 220g, 23-cm-diameter plastic disk from 0 to 1800 rpm in 4.6 seconds, without considering the external forces, is 0.431 Nm.
Explanation:In solving the question,
Torque
is our primary interest. We first need to convert rpm to rad/s since Torque calculations require SI units. The conversion can be done by the formula ω = 2π (frequency), and frequency is simply rpm/60. Hence, 1800 rpm is equivalent to 188.50 rad/s. Now, we use the Kinematics equation ω = ω
0
+ αt to calculate angular acceleration (α), where ω
0
is the initial angular velocity, and it is 0 rad/s in this case as the disk starts from rest, ω is the final angular velocity and is 188.50 rad/s, while t is the time of 4.6 seconds. Solving this gives us α=41 rad/s
2
. The Torque can now be calculated using τ=Iα where I (moment of inertia for a disk) = 0.5*m*r
2
. Substituting the values of m, r and α gives a Torque value of 0.431 Nm.
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The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is assumed constant at 0.0075 lb/ft3 , calculate the elevation of the mountain top.
To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.
For mercury, density, gravity and height are defined as
[tex]\rho_m = 846lb/ft^3[/tex]
[tex]g = 32.17405ft/s^2[/tex]
[tex]h_1 = 1in = \frac{1}{12} ft[/tex]
For the air the defined properties would be
[tex]\rho_a = 0.0075lb/ft^3[/tex]
[tex]g = 32.17405ft/s^2[/tex]
[tex]h_2 = ?[/tex]
We have for equilibrium that
[tex]\text{Pressure change in Air}=\text{Pressure change in Mercury}[/tex]
[tex]\rho_m g h_1 = \rho_a g h_2[/tex]
Replacing,
[tex](846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)[/tex]
Rearranging to find [tex]h_2[/tex]
[tex]h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}[/tex]
[tex]h = 9400ft[/tex]
Therefore the elevation of the mountain top is 9400ft
The elevation of the mountain top is approximately 13,972 feet.
Explanation:The difference in barometric pressure between the sea level and the top of the mountain represents the hydrostatic pressure exerted by the column of air. We can calculate the elevation of the mountain top by using the equation:
ΔP = ρgh
Where ΔP is the difference in pressure, ρ is the density of the air, g is the acceleration due to gravity, and h is the elevation. Rearranging the equation, we have:
h = ΔP / (ρg)
Substituting the given values, we have:
h = (30 - 29) in / (0.0075 lb/ft³ * 32.2 ft/sec²)
Simplifying the equation, we get:
h ≈ 13,972 ft
Therefore, the elevation of the mountain top is approximately 13,972 feet.