Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole
Explanation :
The steps involved in the born-Haber cycle for the formation of [tex]CsCl[/tex] :
(1) Conversion of solid calcium into gaseous cesium atoms.
[tex]Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)[/tex]
[tex]\Delta H_s[/tex] = sublimation energy of calcium
(2) Conversion of gaseous cesium atoms into gaseous cesium ions.
[tex]Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)[/tex]
[tex]\Delta H_I[/tex] = ionization energy of calcium
(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.
[tex]Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)[/tex]
[tex]\Delta H_D[/tex] = dissociation energy of chlorine
(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.
[tex]Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)[/tex]
[tex]\Delta H_E[/tex] = electron affinity energy of chlorine
(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.
[tex]Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)[/tex]
[tex]\Delta H_L[/tex] = lattice energy of calcium chloride
To calculate the overall energy from the born-Haber cycle, the equation used will be:
[tex]\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L[/tex]
Now put all the given values in this equation, we get:
[tex]-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L[/tex]
[tex]\Delta H_L=-667KJ/mole[/tex]
The negative sign indicates that for exothermic reaction, the lattice energy will be negative.
Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole
The magnitude of Lattice energy of CsCl is -676 kJ/mol.
Given Here,
Enthalpy of sublimation of Cs [tex]\rm \bold{ \Delta H(sub ) }[/tex] = +76 kJ/mo
Ionization Energy for Potassium IE(Cs) = +376 kJ/mol
Electron affinity for Chlorine is EA(Cl) = −349 kJ/mol.
Bond dissociation energy of Chlorine, BE(Cl) = +121 kJ/mol
Enthalpy of formation for CsCl, [tex]\rm \bold{ \Delta H(f) }[/tex] = −436.5 kj/mol .
The lattice energy of KCl can be calculated from the formula,
[tex]\rm \bold {U(CsCl) = \Delta Hf(CsCl) - [ \Delta H(sub) + IE(Cs) +BE(Cl_2) + EA(Cl)]}[/tex]
U( CsCl) = -436 - [+76 +376 +121 kJ/mol -349]
U( CsCl) = -676 kJ/mol
Hence we can calculate that the magnitude of Lattice energy of CsCl is -676 kJ/mol.
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Two red blood cells each have a mass of 9.05×10−14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion arising from the excess charge prevents the cells from clumping together. One cell carries −2.50 pC and the other −3.30 pC, and each cell can be modeled as a sphere 3.75×10−6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed, what initial speed would each need so that they get close enough to just barely touch? Assume that there is no viscous drag from any of the surrounding liquid.
Final answer:
The student must use Coulomb's law to calculate the net electrostatic force between two charged red blood cells, and then use energy conservation to determine the initial speed needed for the cells to overcome this force and just touch. So, the initial speed of each cell is approximately [tex]\( 662.8 \, \text{m/s} \)[/tex].
Explanation:
Coulomb's Law and Initial Speed Calculation
The student's problem involving two red blood cells carrying negative charges requires the use of Coulomb's law to determine the electrostatic force of repulsion between them. Since the cells are modeled as spheres, the charges are assumed to be uniformly distributed. To find the initial speed needed for the cells to just touch, one must consider energy conservation principles, where the initial kinetic energy of the cells should be equivalent to the electrostatic potential energy at the point of closest approach (when they just touch).
The formula for Coulomb's law is
F = k * |q1 * q2| / r^2, where F is the force between the charges, q1 and q2 are the charges, r is the separation distance, and k is Coulomb's constant (8.9875 x 10^9 Nm^2/C^2). The potential energy (U) due to electrostatic force is given by U = k * |q1 * q2| / r. By setting the kinetic energy (1/2 * m * v^2) for both cells equal to U, and solving for v, we can find the initial speed of each cell.
To find the initial speed of each cell using the potential energy and kinetic energy, you can follow these steps:
Given:
[tex]- Charges: \( q_1 = -2.50 \, \text{pC} \), \( q_2 = -3.30 \, \text{pC} \)\\- Separation distance: \( r = 3.75 \times 10^{-6} \, \text{m} \)\\- Coulomb's constant: \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)\\- Mass: \( m = 9.05 \times 10^{-14} \, \text{kg} \)\\[/tex]
1. Calculate the potential energy \( U \) using Coulomb's law:
[tex]\[ U = \frac{k \cdot |q_1 \cdot q_2|}{r} \] \[ U = \frac{8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot |(-2.50 \times 10^{-12} \, \text{C}) \cdot (-3.30 \times 10^{-12} \, \text{C})|}{3.75 \times 10^{-6} \, \text{m}} \][/tex]
2. Calculate the potential energy \( U \):
[tex]\[ U = \frac{8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot 8.25 \times 10^{-24} \, \text{C}^2}{3.75 \times 10^{-6} \, \text{m}} \] \[ U = \frac{7.465625 \times 10^{-14} \, \text{Nm}}{3.75 \times 10^{-6} \, \text{m}} \] \[ U = 1.989 \times 10^{-8} \, \text{J} \][/tex]
3. Set the kinetic energy equal to the potential energy:
[tex]\[ \frac{1}{2} m v^2 = U \][/tex]
4. Solve for \( v \):
[tex]\[ v = \sqrt{\frac{2U}{m}} \] \[ v = \sqrt{\frac{2 \times 1.989 \times 10^{-8} \, \text{J}}{9.05 \times 10^{-14} \, \text{kg}}} \][/tex]
5. Calculate \( v \):
[tex]\[ v \approx \sqrt{\frac{2 \times 1.989 \times 10^{-8} \, \text{J}}{9.05 \times 10^{-14} \, \text{kg}}} \] \[ v \approx \sqrt{4.391 \times 10^5 \, \text{m}^2/\text{s}^2} \] \[ v \approx 662.8 \, \text{m/s} \][/tex]
So, the initial speed of each cell is approximately [tex]\( 662.8 \, \text{m/s} \)[/tex].
Since there is no viscous drag, the conservation of energy provides the necessary framework to equate the initial kinetic energy to the electrostatic potential energy. Accounting for the fact that each blood cell has a different charge, the calculation would involve determining the net repulsive force and then the corresponding speed for each cell to overcome this force and just barely touch.
The initial speed required for each red blood cell to just barely touch is approximately [tex]\( 1.479 \times 10^{3} \)[/tex]m/s.
Given:
- Mass of each red blood cell, [tex]\( m = 9.05 \times 10^{-14} \)[/tex] kg
- Charge on the first cell, [tex]\( q_1 = -2.50 \) pC \( = -2.50 \times 10^{-12} \) C[/tex]
- Charge on the second cell, [tex]\( q_2 = -3.30 \) pC \( = -3.30 \times 10^{-12} \) C[/tex]
- Radius of each cell, [tex]\( r = 3.75 \times 10^{-6} \) m[/tex]
When the cells just barely touch, the distance between their centers will be 2r , since each cell has a radius r.
The electrostatic force F between the two charges can be calculated using Coulomb's law:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
where \( k \) is Coulomb's constant, [tex]\( k = 8.9875 \times 10^9 \)[/tex] N m²/C².
Substituting the values, we get:
[tex]\[ F = (8.9875 \times 10^9) \frac{|(-2.50 \times 10^{-12})(-3.30 \times 10^{-12})|}{(2 \times 3.75 \times 10^{-6})^2} \][/tex]
Now, we need to find the initial kinetic energy ( KE ) of each cell, which must be equal to the work done by the electrostatic force when the cells move from infinity to a distance ( 2r ) apart. The work done by the electrostatic force is equal to the potential energy U at a distance 2r:
[tex]\[ U = k \frac{|q_1 q_2|}{2r} \][/tex]
The initial kinetic energy of each cell is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
where v is the initial speed we want to find.
Since there are no other forces doing work (no viscous drag), the total mechanical energy is conserved, and the initial kinetic energy must equal the potential energy at the point of touching:
[tex]\[ \frac{1}{2} m v^2 = k \frac{|q_1 q_2|}{2r} \][/tex]
Solving for v, we get:
[tex]\[ v = \sqrt{\frac{2k |q_1 q_2|}{m (2r)}} \][/tex]
Substituting the values, we have:
[tex]\[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times |(-2.50 \times 10^{-12})(-3.30 \times 10^{-12})|}{9.05 \times 10^{-14} \times (2 \times 3.75 \times 10^{-6})}} \][/tex]
Now, let's calculate the value of \( v \):
[tex]\[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times (2.50 \times 10^{-12} \times 3.30 \times 10^{-12})}{9.05 \times 10^{-14} \times (2 \times 3.75 \times 10^{-6})}} \] \[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times (8.25 \times 10^{-24})}{9.05 \times 10^{-14} \times (7.5 \times 10^{-6})}} \] \[ v = \sqrt{\frac{2 \times (7.4234375 \times 10^{-14})}{6.79875 \times 10^{-20}}} \][/tex]
[tex]\[ v = \sqrt{\frac{14.846875 \times 10^{-14}}{6.79875 \times 10^{-20}}} \] \[ v = \sqrt{2.18375 \times 10^{6}} \] \[ v \approx 1.479 \times 10^{3} \text{ m/s} \][/tex]
Intensity of a wave is the. (a) Power per unit area (b) Power per unit volume (c) Power per unit time (d) All of the above
Answer:
Intensity of a wave is Power per unit area.
Explanation:
The intensity of wave is defined as the power per unit area in the direction of motion of wave i.e.
[tex]I=\dfrac{P}{A}[/tex]
Also, the intensity of a wave is directly proportional to the square of its amplitude. Let a is the amplitude of a wave. So,
[tex]I\propto a^2[/tex]
The SI unit of power is watt and the SI unit of area is m². So, the SI unit of intensity is W/m². Hence, the correct option is (a) "Intensity of a wave is Power per unit area".
In a region where there is a uniform electric field that is upward and has magnitude 3.80x104 N/C a small object is projected upward with an initial speed of 2.32 m/s The object travels upward a distance of 5.98 cm in 0 200 s. What is the object's charge-to-mass ratio q/m (magnitude and sign)?Assume g 9.80 m/s and ignore air resistance E3? C/kg q/m
Answer:
6.03 x 10^-3 C/Kg
Explanation:
E = 3.8 x 10^4 N/C, u = 2.32 m/s, s = 5.98 cm = 0.0598 m, t = 0.2 s, g = 9.8 m/s^2
Acceleration on object is a .
Use second equation of motion.
S = u t + 1/2 a t^2
0.0598 = 2.32 x 0.2 + 0.5 x a x 0.2 x 0.2
0.0598 = 4.64 + 0.02 x a
a = - 229 m/s^2
Now, F = ma = qE
q / m = a / E = 229 / (3.8 x 10000)
q / m = 6.03 x 10^-3 C/Kg
Calculate the charge-to-mass ratio of the object given initial speed, distance, time, and electric field strength.
Given: Electric field magnitude = 3.80 x 10^4 N/C, initial speed = 2.32 m/s, distance traveled = 5.98 cm, time = 0.200 s, acceleration due to gravity = 9.80 m/s^2.
To find: The object's charge-to-mass ratio (q/m).
Calculations: Determine the acceleration using the provided data, then relate the forces acting on the object (gravity and electric field) to find q/m.
A uniform solid sphere of radius r = 0.490 m and mass m = 13.5 kg turns counterclockwise about a vertical axis through its center (when viewed from above). Find its vector angular momentum about this axis when its angular speed is 2.9 rad/s.
Answer:
Vector angular momentum about this axis of the sphere is:
L= 3.76[tex]\hat k[/tex] kg-m²/sec
Explanation:
The formula for the moment of inertia of a sphere is:
[tex]I=\frac{2}{5}\times MR^2[/tex]
Given:
Mass of the sphere = 13.5 kg
Radius of the sphere = 0.490 m
Thus, moment of inertia :
[tex]I=\frac{2}{5}\times 13.5\times (0.490)^2 kg\ m^2[/tex]
[tex]I=1.29654 kg\ m^2[/tex]
The expression for the angular momentum is:
L=I×ω
Given:
Angular speed(ω) = 2.9 rad/s
I, above calculated = 1.29654 kgm⁻²
Thus, angular momentum is:
L= 1.29654×2.9 kg-m²/sec
L= 3.76 kg-m²/sec
Given, the sphere is turning counterclockwise about the vertical axis. Thus, the direction of the angular momentum will be on the upper side of the plane. ( [tex]+\hat k[/tex] ).
Thus, angular momentum with direction is:
L= 3.76[tex]\hat k[/tex] kg-m²/sec
1. Assume that the dunes in a field of barchans sand dunes are moving at an average long-term rate of 30 meters (about 100 feet) a year. How many years does it take a dune to move a distance of 500 meters?
Answer:
16.67 years
Explanation:
It moves a distance of 30 m in a year.
The time required tp move 500 m = 500 / 30 = 16.67 years
The total electromagnetic power emitted by the sun is 3.8 × 1026 W. What is the radiation pressure on a totally absorbing satellite at the orbit of Mercury, which has an orbital radius of 5.8 × 1010 m?
Answer:
[tex]3.0\cdot 10^{-5} N/m^2[/tex]
Explanation:
The intensity of the radiation at the location of the satellite is given by:
[tex]I=\frac{P}{4\pi r^2}[/tex]
where
[tex]P=3.8\cdot 10^{26}W[/tex] is the power of the emitted radiation
[tex]4\pi r^2[/tex] is the area over which the radiation is emitted (the surface of a sphere), with
[tex]r=5.8\cdot 10^{10}m[/tex] being the radius of the orbit
Substituting,
[tex]I=\frac{3.8\cdot 10^{26}}{4\pi (5.8\cdot 10^{10})^2}=8989 W/m^2[/tex]
Now we can find the pressure of radiation, that for a totally absorbing surface (such as the satellite) is:
[tex]p=\frac{I}{c}[/tex]
where
[tex]c=3.0\cdot 10^8 m/s[/tex]
is the speed of light. Substituting,
[tex]p=\frac{8989}{3.0\cdot 10^8}=3.0\cdot 10^{-5} N/m^2[/tex]
A pump lifts water from a lake to a large tank 20 m above the lake. How much work against gravity does the pump do as it transfers 5.0 m^3 if the density is ?=1000kg/m^3?
Answer:
980 kJ
Explanation:
Work = change in energy
W = mgh
W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)
W = 980,000 J
W = 980 kJ
The pump does 980 kJ of work.
The work done by the pump against gravity in transfering 5 m³ of water to a tank 20 m above the lake is 980000 J
We'll begin by calculating the mass of the water. This can be obtained as follow:
Volume of water = 5 m³
Density of water = 1000 Kg/m³
Mass of water =?Mass = Density × Volume
Mass of water = 1000 × 5
Mass of water = 5000 KgFinally, we shall determine the work done by the pump. This can be obtained as illustrated below:Mass of water (m) = 5000 Kg
Height (h) = 20 m
Acceleration due to gravity (g) = 9.8 m/s²
Work done = ?Work done = mgh
Work done = 5000 × 9.8 × 20
Work done = 980000 JThus, the work done by the pump is 980000 J
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A planet is 15 light years from earth. At which of the following speeds will the crew of a spaceship complete the trip to the planet in 8 years?
A) 0.78c
B) 0.85c
C) 0.88c
D) 0.92c
Answer:
option c) 0.88c is the correct answer
Explanation:
using the Lorrentz equation we have
[tex]t=\frac{d}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]
where,
t = time taken to cover the distance
d = Distance
v = velocity
c = speed of light
given
d = 15 light years
Now,
[tex]8years=\frac{15years\times c}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]
or
[tex](\frac{v}{c})^2\times (\frac{8}{15})^2=1-(\frac{v}{c})^2[/tex]
or
[tex](\frac{v}{c})^2\times 0.284=1-(\frac{v}{c})^2[/tex]
or
[tex](\frac{v}{c})^2\times 0.284 + (\frac{v}{c})^2 =1[/tex]
or
[tex](1.284\times \frac{v}{c})^2 =1[/tex]
or
[tex] (\frac{v}{c}) =\sqrt{\frac{1}{0.284}}[/tex]
or
[tex]v=0.88c[/tex]
A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is the roof? Note: the acceleration of an object due to earth's gravity is 32 f t sec 2
Answer:
d = 506.25 ft
Explanation:
As we know by kinematics that
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we know that initially the stone is dropped from rest from the edge of the roof
so here initial speed will be zero
now we have
[tex]v_i = 0[/tex]
also the acceleration of the stone is due to gravity which is given as
[tex]g = 32 ft/s^2[/tex]
now we have
[tex]v_f = 180 ft/s[/tex]
so from above equation
[tex]180^2 - 0 = 2(32)d[/tex]
[tex]d = 506.25 ft[/tex]
Using the equation of motion, we find that the height of the roof from which the stone was dropped is approximately 506.25 feet.
Explanation:The subject question involves a physics concept relating to the motion of objects under the influence of gravity, in particular, the calculation of displacement during free fall. We can solve this problem using one of the fundamental equations of motion: final velocity2 = initial velocity2 + 2*a*d. In this case, initial velocity (u) = 0 (as the stone was dropped), final velocity (v) = -180 feet per second (negative as the stone is moving downwards), and acceleration (a) due to gravity is -32 feet/second2 (negative as gravity acts downward).
Substituting these values, we get:
(-180)2 = (0)2 + 2*(-32)*d.
This simplifies to:
32400 = -64d.
Solving for d (which represents the height of the roof), we get:
d = -32400 / -64 = 506.25 feet. Hence, the roof is approximately 506.25 feet high.
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A block weighing 87.0 N rests on a plane inclined at 24.1° to the horizontal. The coefficient of the static and kinetic frictions are 0.25 and 0.13 respectively. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the block from slipping?
Answer:
15.7 N
Explanation:
Draw a free body diagram. The block has four forces acting on it. Gravity pulling down, normal force pushing perpendicular to the plane, friction pointing up the plane, and applied force F pushing up the plane.
Sum of the forces normal to the plane:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the plane:
∑F = ma
Nμ + F − mg sin θ = 0
F = mg sin θ − Nμ
Substituting:
F = mg sin θ − mgμ cos θ
F = mg (sin θ − μ cos θ)
Given mg = 87.0 N, θ = 24.1°, and μ = 0.25 (because the block is not moving):
F = 87.0 N (sin 24.1° − 0.25 cos 24.1°)
F = 15.7 N
The minimum magnitude of the force F required to prevent the block from slipping is approximately 14.114 N.
The minimum magnitude of the force F required to prevent the block from slipping is given by the equation:
[tex]\[ F = w \sin(\theta) - \mu_s w \cos(\theta) \][/tex]
where w is the weight of the block, [tex]\( \theta \)[/tex] is the angle of inclination, and [tex]\mu_s \)[/tex] is the coefficient of static friction.
Given:
-[tex]\( w = 87.0 \, \text{N} \)[/tex]
- [tex]\( \theta = 24.1^\circ \)[/tex]
- [tex]\( \mu_s = 0.25 \)[/tex]
First, convert the angle from degrees to radians because the sine and cosine functions in most calculators require the input to be in radians:
[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180^\circ} \][/tex]
[tex]\[ \theta_{\text{rad}} = 24.1^\circ \times \frac{\pi}{180^\circ} \approx 0.420 \text{ radians} \][/tex]
Now, calculate the minimum force F using the given formula:
[tex]\[ F = w \sin(\theta_{\text{rad}}) - \mu_s w \cos(\theta_{\text{rad}}) \][/tex]
[tex]\[ F = 87.0 \, \text{N} \times \sin(0.420) - 0.25 \times 87.0 \, \text{N} \times \cos(0.420) \][/tex]
[tex]\[ F \ =87.0 \, \text{N} \times 0.412 - 0.25 \times 87.0 \, \text{N} \times 0.910 \][/tex]
[tex]\[ F \ = 35.864 \, \text{N} - 21.75 \, \text{N} \][/tex]
[tex]\[ F \ =14.114 \, \text{N} \][/tex]
A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of work on the hoop. What is the new speed of the center of mass of the hoop (in m/s)? Round your answer to the nearest whole number.
Answer:
[tex]v_f = 20 m/s[/tex]
Explanation:
Since the hoop is rolling on the floor so its total kinetic energy is given as
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2[/tex]
now for pure rolling condition we will have
[tex]v = R\omega[/tex]
also we have
[tex]I = mR^2[/tex]
now we will have
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}[/tex]
[tex]KE = mv^2[/tex]
now by work energy theorem we can say
[tex]W = KE_f - KE_i[/tex]
[tex]842 J = mv_f^2 - mv_i^2[/tex]
[tex]842 = 3(v_f^2) - 3\times 11^2[/tex]
now solve for final speed
[tex]v_f = 20 m/s[/tex]
Convert the value 43.28 lbf/in3 to standard SI units
Answer:
[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]
Explanation:
Here the given values corresponds to the density of a substance
The density of a substance in standard SI unit is given as Kg/m³.
Given value = 43.28lbf/in³
also
we know,
1 lbf = 0.45kg
1in = 0.0254 m
thus,
[tex]\frac{1lbf}{1in^3}=\frac{0.45kg}{0.0254^3m^3}=27460.68\frac{kg}{m^3}[/tex]
therefore,
[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]
What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter is 0.50 mm?
R = 0.407Ω.
The resistance R of a particular conductor is related to the resistivity ρ of the material by the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of the material.
To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.
We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4. Then:
R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]
R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²
R = 0.407Ω
To determine the resistance of this wire is equal to 0.4082 Ohms.
Given the following data:
Resistivity = [tex]3.2 \times 10^{-8}[/tex] Ω.m.Length = 2.5 mDiameter = 0.50 mm to m = [tex]5\times 10^{-4}[/tex]To determine the resistance of this wire:
How to calculate the resistance.First of all, we would determine the area of the wire by using this formula:
[tex]A =\frac{\pi d^2}{4} \\\\A =\frac{3.142 \times (5\times 10^{-4})^2}{4}\\\\A = 1.96 \times 10^{-7}\;m^2[/tex]
Now, we can determine the resistance of this wire:
[tex]R=\frac{\rho L}{A} \\\\R=\frac{3.2 \times 10^{-8} \times 2.5}{1.96 \times 10^{-7}}[/tex]
Resistance = 0.4082 Ohms.
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An electron is travelling East through a magnetic field directed South. In which direction is the force on the electron?
Answer:
Vertically upwards
Explanation:
Fleming's left hand rule: it states taht when we spread our fore finger, middle finger and thumb in such a way that they are mutually perpendicular to each other, theN thumb indicates the direction of force, fore finger indicates the direction of magnetic field and the middle finger indicates the direction of current.
So according to this rule the direction of force is vertically upwards.
Frequency and amplitude of a particle in simple harmonic motion A particle moves in simple harmonic motion. Knowing that the maximum velocity is 200 mm/s and the maximum acceleration is 13 m/s , determine the amplitude and frequency of the motion. Amplitude of the motion xm 1mm. Frequency of the motion勿 Hz.
Answer:
amplitude, a = 3.076 mm
frequency, f = 10.35 Hz
Explanation:
Vmax = 200 mm /s = 0.2 m/s
Amax = 13 m/s
Let f be the frequency and a be the amplitude.
Use the formula of maximum velocity.
Vmax = ω a
0.2 = ω a ..... (1)
Use the formula of maximum acceleration.
Amax = ω^2 x a
13 = ω^2 a ..... (2)
Divide equation (2) by (1)
ω = 13 / 0.2 = 65 rad/s
Put in equation (1)
0.2 = 65 x a
a = 3.076 x 10^-3 m
a = 3.076 mm
Let f be the frequency
ω = 2 π f
f = 65 / (2 x 3.14) = 10.35 Hz
Please show all your work
After an initially neutral glass rod is rubbed with an initially neutral silk scarf, the rod has a charge of 89.0 µC. Estimate the fractional increase or decrease in the scarf's mass. (Assume the scarf had a mass of 100 g.)
delta M / M = ?
Would this change in mass be easily noticed? (Assume that this change is noticeable if it is more than 1 mg.)
yes or no
Answer:
5.062 x 10^-9 mg
Explanation:
Charge on glass rod = 89 uC
Charge on silk cloth = - 89 uC
mass of scarf, M = 100 g
Number of excess electrons on silk cloth = charge / charge of one electron
n = (89 x 10^-6) / ( 1.6 x 10^-19)
n = 5.5625 x 10^14
mass of one electron = 9.1 x 10^-31 kg
Mass of 5.5625 x 10^14 electrons = 5.5625 x 10^14 x 9.1 x 10^-31 kg
ΔM = 5.062 x 10^-16 kg
Δ M / M = (5.062 x 10^-16) / 0.1 = 5.062 x 10^-15 kg = 5.062 x 10^-9 mg
It is not noticeable.
A 0.0811 kg ice cube at 0 °C is dropped into a Styrofoam cup holding 0.397 kg of water at 14.8 °C. Calculate the final temperature of the system. Assume the cup and the surroundings can be ignored.
Final answer:
The question revolves around the final temperature of a mixture of ice and water in a thermally insulated system. The specific heats and enthalpy of fusion are required to solve, but the proper method involves using the conservation of energy.
Explanation:
The student is asking about the final temperature of a system consisting of a 0.0811 kg ice cube at 0°C and 0.397 kg of water at 14.8°C. To solve this, we can use the principle of conservation of energy, stating that the heat lost by the water will equal the heat gained by the ice cube as it warms up, melts, and then possibly continues to heat up as water. However, without knowing the specific heat capacities and the enthalpy of fusion, we cannot provide a numerical answer. Given the specific heats and the heat of fusion, one would set up an equation where the heat lost by the water equals the heat gained by the ice, and solve for the final temperature. The specific heat of water, the specific heat of ice, and the enthalpy of fusion of ice would be necessary information to perform actual calculations.
Find the acceleration due to Earth’s gravity at the distance of the Moon, which is on average 3.84 × 10^8 m from the center of Earth.
Answer:
[tex]a = 2.7 \times 10^{-3} m/s^2[/tex]
Explanation:
As we know that acceleration due to gravity is given as
[tex]a = \frac{GM}{r^2}[/tex]
here we know that
[tex]M = 5.97 \times 10^{24} kg[/tex] (mass of earth)
r = distance at which acceleration is to be find out
[tex]r = 3.84 \times 10^8 m[/tex]
now from the formula of acceleration due to gravity we know
[tex]a = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(3.84 \times 10^8)^2}[/tex]
[tex]a = 2.7 \times 10^{-3} m/s^2[/tex]
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Answer:
YES THERE ARE VERY MANY BOOKS THAT YOU CAN BUY OR BORROW FROM YOUR LOCAL LIBRARY, HOPE THIS HELPS! HAVE A GREAT DAY!
Explanation:
A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm long. Next you pull down on the weight so the spring is 25.2 cm long and you release the weight from rest. What is the period of oscillation?
Answer:
Time period of oscillations is 0.62 s
Explanation:
Due to suspension of weight the change in the length of the spring is given as
[tex]\Delta L = L_f - L_i[/tex]
[tex]\Delta L = 21.5 - 11.9 = 9.6 cm[/tex]
now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force
[tex]mg = kx[/tex]
[tex]0.037 (9.81) = k(0.096)[/tex]
[tex]k = 3.78 N/m[/tex]
Now the period of oscillation of spring is given as
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
Now plug in all values in it
[tex]T = 2\pi \sqrt{\frac{0.037}{3.78}}[/tex]
[tex]T = 0.62 s[/tex]
The period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.
What is the value of the spring constant?We know that the formula of the spring force is written as,
[tex]F = k \times \delta[/tex]
Given to us,
weight on the spring, F = 37 g = (9.81 x 0.037) = 0.36297 N
Deflection in spring, δ = (21.5 - 11.9) = 9.6 cm = 0.096 m
Substitute the value,
[tex]0.36297 = k \times 0.096\\k= 3.78\rm\ N/m[/tex]
What is the force that is been applied to the spring?We know that the weight is been pulled down therefore, an external force is been applied to the spring to extend it further to 25.2 cm.
We know that the formula of the spring force is written as,
[tex]F = k \times \delta[/tex]
Given to us
Deflection in the spring, δ = (25.2 - 11.9) = 13.3 cm = 0.133 m
Spring constant, k = 3.78 N/m
Substitute the values,
[tex]F = 3.78 \times 0.133\\\\F = 0.5\\\\m \times g = 0.5\\\\m = 0.05125\rm\ kg[/tex]
What is the period of oscillation?We know the formula for the period of oscillation is given as,
[tex]T = 2\pi \sqrt{\dfrac{m}{k}}[/tex]
Substitute the values,
m = 0.05125 kg
k = 3.78 N/m
[tex]T = 2\pi \sqrt{\dfrac{0.05125}{3.78}}\\T = 0.7316\rm\ sec[/tex]
Hence, the period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.
Learn more about Oscillation:
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Given the time of flight for a projectile fired horizontally was o.66 seconds, and it landed 3:3 m from where it was launched, the projectile's muzzle velocity was: (A) 0.20 m/s, (B) 0.22 m/s, (C) 0.50 m/s, (D) 5.0 m/s, (E) none of the above.
Answer:
Option (D)
Explanation:
In the projectile motion
Horizontal distance = horizontal velocity × time
3.3 = u × 0.66
u = 5 m/s
Muzzle velocity is 5 m/s
An athlete swings a 4.00 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.860 m at an angular speed of 0.660 rev/s. (a) What is the tangential speed of the ball? m/s m/s2 m/s (b) What is its centripetal acceleration? (c) If the maximum tension the rope can withstand before breaking is 136 N, what is the maximum tangential speed the ball can have?
Answer:
a) Tangential speed = 3.57 m/s
b) Centripetal acceleration = 17.79 m/s²
c) Maximum tangential speed the ball can have = 5.41 m/s
Explanation:
a) Tangential speed, v = rω
Radius, r = 0.86 m
Angular speed, ω = 0.660 rev/s = 0.66 x 2 x π = 4.15 rad/s
Tangential speed, v = rω = 0.86 x 4.15 = 3.57 m/s
b) Centripetal acceleration,
[tex]a=\frac{v^2}{r}=\frac{3.57^2}{0.86}=14.79m/s^2[/tex]
c) Maximum tension in horizontal circular motion
[tex]T=\frac{mv^2}{r}\\\\136=\frac{4\times v^2}{0.86}\\\\v=5.41m/s[/tex]
Maximum tangential speed the ball can have = 5.41 m/s
The questions involve concepts of circular motion: tangential speed, centripetal acceleration, and the tension in the rope maintaining the motion. Tangential speed is derived from the radius of the circle and the angular velocity, centripetal acceleration is derived from the square of tangential speed divided by the radius, and the maximum tangibility speed before the rope breaks can be calculated from its tension.
Explanation:This is a problem of circular motion, and it's crucial to understand the concepts of tangential speed, centripetal acceleration, and the tension in the rope.
The tangential speed of a body moving along a circular path is its normal linear speed for that moment. It can be calculated by using the formula v = rω, where v is the tangential speed, r is the radius of the circle, and ω is the angular velocity. Make sure to convert the angular speed from rev/s to rad/s (ω = 0.660 rev/s * 2π rad/rev).
Centripetal acceleration is the rate of change of tangential velocity, and it always points toward the center of the circle. It can be calculated using the formula ac = v²/r.
The tension in the rope is the force exerting on the ball to keep it moving circularly. If it exceeds its maximum limit (136N in this case), the rope will break. The maximum tangential speed can be derived from the equation T = m * v² / r.
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A motorcycle is moving at 15 m/s when its brakes are applied, bringing the cycle to rest in 5.3 s. To the nearest meter, how far does the motorcycle travel while coming to a stop?
Answer:
Motorcycle travel 39.75 m while coming to a stop
Explanation:
We have equation of motion
v = u + at
Final velocity, v = 0 m/s, u = 15 m/s, t = 5.3 s
0 = 15 + a x 5.3
a = -2.83 m/s²
Now we have the other equation of motion
v² = u² + 2as
0² = 15² - 2 x 2.83 x s
s = 39.75 m
Motorcycle travel 39.75 m while coming to a stop
Three resistors, R1 = 10 Ω, R2 = 20 Ω and R3 = 30 Ω are connected in series to a 12 V battery. Find: a. The equivalent resistance of the circuit. b. The current in each resistor. c. The voltage across each resistor. d. The power lost in each resistor.
Explanation:
a) The equivalent resistance for resistors in series is the sum:
R = R₁ + R₂ + R₃
R = 10 Ω + 20 Ω + 30 Ω
R = 60 Ω
b) The resistors are in series, so they have the same current.
I = V/R
I = 12 V / 60 Ω
I = 0.2 A
c) The voltage drop can be found with Ohm's law:
V₁ = I R₁ = (0.2 A) (10 Ω) = 2 V
V₂ = I R₂ = (0.2 A) (20 Ω) = 4 V
V₃ = I R₃ = (0.2 A) (30 Ω) = 6 V
d) The power lost in each resistor is current times voltage drop:
P₁ = I V₁ = (0.2 A) (2 V) = 0.4 W
P₂ = I V₂ = (0.2 A) (4 V) = 0.8 W
P₃ = I V₃ = (0.2 A) (6 V) = 1.2 W
(a) The equivalent resistance of the circuit is 60 Ω.
(b) The current in each resistor is 0.2 A.
(c) The voltage across each resistor is 2 V, 4 V, and 6 V respectively.
(d) The power lost in each resistor is 0.4 W, 0.8 W and 1.2 W.
Equivalent resistance
The equivalent resistance of the circuit is determined as follows;
Rt = R₁ + R₂ + R₃
Rt = 10 + 20 + 30
Rt = 60 Ω
Current in the resistorThe current in the resistors is calculated as follows;
V = IR
I = V/R
I = 12/60
I = 0.2 A
Voltage across each resistorV1 =IR1 = 0.2 x 10 = 2 V
V2 = IR2 = 0.2 x 20 = 4 V
V3 = 0.2 x 30 = 6 V
Power lost in each resistorP1 = I²R1 = (0.2)² x 10 = 0.4 W
P2 = I²R2 = (0.2)² x 20 = 0.8 W
P3 = I²R3 = (0.2)² x 30 = 1.2 W
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A tightly wound solenoid of 1600 turns, cross-sectional area of 6.00 cm2, and length of 20.0 cm carries a current of 2.80 A. (a) What is its inductance? (b) If the cross-sectional area is doubled, does anything happen to the value of the inductance? Explain your answer.
Answer:
(a) 9.65 mH
(b) doubled
Explanation:
A = 6 cm^2 = 6 x 10^-4 m^2, l = 20 cm = 0.2 m, i = 2.8 A, N = 1600
(a) The formula for the self inductance of a solenoid is given by
L = μ0 N^2 A / l
L = 4 x 3.14 x 10^-7 x 1600 x 1600 x 6 x 10^-4 / 0.2
L = 9.65 x 10^-3 H
L = 9.65 mH
(b) As we observe that the inductance is proportional to the area so if the area is doubled, then inductance is also doubled.
What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength in angstroms (x 10-10 m) to 2 sig figs.
Answer:
1.2826 x 10^-13 m
Explanation:
[tex]\lambda = \frac{h}{\sqrt{2 m K}}[/tex]
Here, k be the kinetic energy and m be the mass
K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J
m = 1.67 x 10^-27 kg
[tex]\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}[/tex]
λ = 1.2826 x 10^-13 m
A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in the block makes an 27.7° with the normal. What is the index of refraction of the glass?
Final answer:
To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media.
Explanation:
To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media:
n1 * sin(θ1) = n2 * sin(θ2)
Given that the incident angle is 50.0° and the refracted angle is 27.7°, we can plug in the values to find the index of refraction:
n1 * sin(50.0°) = n2 * sin(27.7°)
The index of refraction of the glass is the ratio of the index of refraction of the glass to the index of refraction of air, which is approximately 1:
n2 / n1 = 1 / sin(27.7°)
A piece of cork (density 250 kg/m3 ) of mass 0.01 kg is held in place under water (density 1000 kg/m3 ) by a string. What is the tension, T, in the string? [Use g = 10 m/s2 ]
Answer:
0.3 N
Explanation:
mass of cork = 0.01 kg, density of cork = 250 kg/m^3
density of water = 1000 kg/m^3, g = 10 m/s^2
Tension in the rope = Buoyant force acting on the cork - Weight of the cork
Buoyant force = volume of cork x density of water x g
= mass x density of water x g / density of cork
= 0.01 x 1000 x 10 / 250 = 0.4 N
Weight of cork = mass of cork x g = 0.01 x 10 = 0.1 N
Thus, the tension in the rope = 0.4 - 0.1 = 0.3 N
Answer:
Tension = 0.3 N
Explanation:
As we know that the cork is inside water
so the buoyancy force on the cork is counter balanced by tension force in string and weight of the block
So the force equation is given as
[tex]F_b = T + mg[/tex]
now we will have
[tex]Volume = \frac{mass}{density}[/tex]
[tex]V = \frac{0.01}{250} = 4 \times 10^{-5} m^3[/tex]
now buoyancy force on the block is given by
[tex]F_b = \rho V g[/tex]
[tex]F_b = 1000(4 \times 10^{-5})(10)[/tex]
[tex]F_b = 0.4 N[/tex]
now by force balance equation
[tex]0.4 = T + 0.01(10)[/tex]
[tex]T = 0.4 - 0.1 = 0.3 N[/tex]
At the equator, near the surface of Earth, the magnetic field is approximately 80.0 μT northward and the electric field is about 150 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron with instantaneous velocity of 6.00 106 m/s directed to the east in this environment.
Answer:
Gravitational force = 89.18 x 10^-31 N
Electric force = 2.4 x 10^-17 N
Magnetic force = 7.68 x 10^-17 N
Explanation:
B = 80 micro tesla = 80 x 10^-6 T north
E = 150 N/C downward
v = 6 x 10^6 m/s east
Gravitational force = m g = 9.1 x 10^-31 x 9.8 = 89.18 x 10^-31 N
Electric force = q E = 1.6 x 10^-19 x 150 = 2.4 x 10^-17 N
Magnetic force = q v B Sin 90 = 1.6 x 10^-19 x 6 x 10^6 x 80 x 10^-6
= 7.68 x 10^-17 N
By how much is the energy stored in a Hooke's law spring increased when its stretch is increased from 7.00 cm to 15.00 cm? a) 159% b) 259% c) 359% d) 459%?
Answer:
c) 359%
Explanation:
k = spring constant of the spring
x₁ = initial stretch of the spring = 7 cm = 0.07 m
x₂ = final stretch of the spring = 15 cm = 0.15 m
Percentage increase in energy is given as
[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{((0.5)k{x_{2}}^{2} - {(0.5)k x_{1}}^{2})(100)}{(0.5)k{x_{1}}^{2}}[/tex]
[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({x_{2}}^{2} - {x_{1}}^{2})(100)}{{x_{1}}^{2}}[/tex]
[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({0.15}^{2} - {0.07}^{2})(100)}{{0.07}^{2}}[/tex]
[tex]\frac{\Delta U \times 100}{U_{1}}[/tex] = 359%