Answer:
D
Sodium atoms donate electrons.
Explanation:
An ionic bond is formed between sodium and chlorine when sodium atoms donate electrons.
What is an ionic bond?Ionic bond or electrovalent bond is a type of bond which is formed between two elements when there is an exchange of electrons which takes place between the atoms resulting in the formation of ions.
When the atom looses an electron it develops a positive charge and forms an ion called the cation while the other atom gains the electron and develops a negative charge and forms an ion called the anion.
As the two atoms are oppositely charged they attract each other which results in the formation of a bond called the ionic bond.Compounds having ionic bonds are called ionic compounds and are hard and have a definite crystalline structure with a high melting point.
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A 150m race is run on a 300m circular track of circumference. Runners start running from the north and turn west until reaching the south. What is the magnitude of the displacement made by the runners?
Answer:
95.5 m
Explanation:
The displacement is the position of the ending point relative to the starting point.
In this case, the magnitude of the displacement is the diameter of the circular track.
d = 300 m / π
d ≈ 95.5 m
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s
Answer:
389 kg
Explanation:
The computation of mass is shown below:-
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
Where K indicates spring constant
m indicates mass
For the new time period
[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]
Now, we will take 2 ratios of the time period
[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]
[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]m' = \frac{0.500}{0.5625}[/tex]
= 0.889 kg
Since mass to be sum that is
= 0.889 - 0.500
0.389 kg
or
= 389 kg
Therefore for computing the mass we simply applied the above formula.
The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.
Given :
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.
The formula of the time period is given by:
[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex] ---- (1)
where m is the mass and K is the spring constant.
The new time period is given by:
[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex] ---- (2)
where m' is the total mass after the addition and K is the spring constant.
Now, divide equation (1) by equation (2).
[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]
Now, substitute the known terms in the above expression.
[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]
Simplify the above expression in order to determine the value of m'.
[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]
m' = 0.889 Kg
Now, the mass added to the object to change the period to 2.00 s is given by:
m" = 0.889 - 0.500
m" = 0.389 Kg
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A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic energy is translational kinetic energy
Answer:
41.667 per cent of the total kinetic energy is translational kinetic energy.
Explanation:
As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:
[tex]E = K_{t} + K_{r}[/tex]
Where:
[tex]E[/tex] - Total energy, measured in joules.
[tex]K_{r}[/tex] - Rotational kinetic energy, measured in joules.
[tex]K_{t}[/tex] - Translational kinetic energy, measured in joules.
The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:
Rotational kinetic energy
[tex]K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}[/tex]
Translational kinetic energy
[tex]K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}[/tex]
Where:
[tex]I_{g}[/tex] - Moment of inertia of the spherical shell with respect to its center of mass, measured in [tex]kg\cdot m^{2}[/tex].
[tex]\omega[/tex] - Angular speed of the spherical shell, measured in radians per second.
[tex]R[/tex] - Radius of the spherical shell, measured in meters.
After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:
[tex]E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}[/tex]
In addition, the moment of inertia of a spherical shell is equal to:
[tex]I_{g} = \frac{2}{3}\cdot m\cdot R^{2}[/tex]
Then, total energy is reduced to this expression:
[tex]E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}[/tex]
The fraction of the total kinetic energy that is translational in percentage is given by the following expression:
[tex]\%K_{t} = \frac{K_{t}}{E}\times 100\,\%[/tex]
[tex]\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%[/tex]
[tex]\%K_{t} = \frac{5}{12}\times 100\,\%[/tex]
[tex]\%K_{t} = 41.667\,\%[/tex]
41.667 per cent of the total kinetic energy is translational kinetic energy.
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each
Answer:
The two masses are 3.39 Kg and 1.75 Kg
Explanation:
The gravitational force of attraction between two bodies is given by the formula;
F = Gm₁m₂/d²
where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²
m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects
Further calculations are provided in the attachment below
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
time =
Answer:
The waiting time is [tex]t_w = 3 .47 \ s[/tex]
Explanation:
From the question we are told that
The height of the hot air balloon above the ground is [tex]d = 50 \ m[/tex]
The distance of the balloon from the target is [tex]l = 100 \ m[/tex]
The velocity of the balloon is [tex]v = 15 \ m/s[/tex]
Generally the time it will take to reach the ground is
[tex]t = \sqrt{2 * \frac{d}{g} }[/tex]
substituting values
[tex]t = \sqrt{2 * \frac{50}{9.8} }[/tex]
[tex]t = 3.2 \ s[/tex]
The distance that is covered at time with the given velocity is mathematically evaluated as
[tex]z = v * t[/tex]
substituting values
[tex]z = 15 * 3.2[/tex]
[tex]z = 48 \ m[/tex]
This implies that for the balloon moving at a velocity (v) to hit the target it must be dropped at this distance (z)
Now the distance the balloonist has to wait before dropping in order to hit the target is
[tex]A = d - z[/tex]
substituting values
[tex]A = 100 - 48[/tex]
[tex]A = 52 \ m[/tex]
This implies that the time the balloonist has to wait is
[tex]t_w = \frac{A}{v}[/tex]
substituting values
[tex]t_w = \frac{52}{15}[/tex]
[tex]t_w = 3 .47 \ s[/tex]
an airplane starts at rest on a runway. it accelerates at 4.44 m/s^2 and travels 738 m before taking off . how long did it take?
Answer:
18.2 s
Explanation:
Given:
Δx = 738 m
v₀ = 0 m/s
a = 4.44 m/s²
Find: t
Δx = v₀ t + ½ at²
738 m = (0 m/s) t + ½ (4.44 m/s²) t²
t = 18.2 s
the density of gold is 19 300kg/m^3. what is the mass of gold cube with the length 0.2015m?
Answer:
The mass is [tex]157.87m^3[/tex]Explanation:
Given data
length of cube= 0.2015 m
density = 19300 kg/m^3.
But the volume of cube is given as [tex]l*l*l= l^3[/tex]
[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]
The density is expressed as = mass/volume
[tex]mass=19300*0.00818= 157.87m^3[/tex]
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
Objects A and B are both positively charged. Both have a mass of 900 g, but A has twice the charge of B. When A and B are placed 30.0 cm apart, B experiences an electric force of 0.870 N.
How large is the force on A?
What is the charge on qA and qB?
If the objects are released, what is the initial acceleration of A?
Answer:
- Force on A = 0.870N
- charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- a = 0.966 m/s^2
Explanation:
- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.
Then, the force on A is 0.870N
- In order to calculate the charge of both objects, you use the following formula:
[tex]F_e=k\frac{q_Aq_B}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
r: distance between the objects = 30.0cm = 0.30m
A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.
You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.
[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]
Then, you have:
charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:
[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]
m: mass of the object A = 900g = 0.900kg
[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]
The acceleration of A is 0.966m/s^2
how do a proton and neutron compare?
Answer:
c.they have opposite charges.
Explanation:
because the protons have a positive charge and the neutrons have no charge.
Fill a Styrofoam cup with very hot water ( or Coffee). Take objects made of different materials, such as a metal spoon, a wooden pencil, a plastic pen or a glass rod. Put one end of each object into the water and answer following questions.
Required:
a. How close to the surface of the water do you have to hold the object for it to feel noticeably warm to the touch?
b. From observations, which material would you judge to be the best conductor of heat? Which is the worst?
c. Rank the materials according to their ability to conduct heat from best to worst.
Answer:
Explanation:
a ) A good conducting material conducts heat easily from one point to another . So the metal spoon which is good conductor will become hot easily by conducting heat from hot water to tip of the spoon . So even after touching the far end of spoon , we can feel the heat of the hot water in the cup . For other material like wood or plastic we have to hold the part which is very close to water to feel heat of water .
b ) The best conductor will be that which touches hot even when we touch its farthest end . Hence metal spoon will be best conductor . Plastic pen will be worst conductor because even touching its part close to hot water , we do not feel much heat .
c ) metal spoon > glass rod > wooden pencil > plastic pen .
1. As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty
Answer:
It's more likely that the trailer is heavily loaded
Explanation:
Due to the fact that the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is very likely that the truck would be heavily loaded because the force constant would be the same whether the truck is empty or heavily loaded.
If a plane is flying level at 910 km/h and the banking angle is not to exceed 50 ∘, what's the minimum curvature radius for the turn?
Answer:
5.5 km
Explanation:
First, we convert the distance from km/h to m/s
910 * 1000/3600
= 252.78 m/s
Now, we use the formula v²/r = gtanθ to get our needed radius
making r the subject of the formula, we have
r = v²/gtanθ, where
r = radius of curvature needed
g = acceleration due to gravity
θ = angle of banking
r = 252.78² / (9.8 * tan 50)
r = 63897.73 / (9.8 * 1.19)
r = 63897.73 / 11.662
r = 5479 m or 5.5 km
Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
what is the value of the tropic of cancer
Answer: The latitudinal value of tropic of cancer is 23.5° N on June 21, when the sun is directly up above the head at noon. The Equator is the circle at which sun is straight above the head.
Explanation:
The rock and meterstick balance at the 25-cm mark, as shown in the sketch. The meterstick has a mass of 1 kg. What must be the mass of the rock? (Show work).
Answer:
1 kgExplanation:
Check the diagram attached below for the diagram.
Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark
Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.
Moment = Force * perpendicular distance
The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction
Clockwise moment = 1kg * 25 = 25kg/cm
Anticlockwise moment = W * 25cm = 25W kg/cm
Equating both moments of forces
25W = 25
W = 25/23
W = 1 kg
The mass of the rock is also 1 kg
________ can occur when an analog connection creates an electromagnetic field around its conductors, inducing its waveforms on a nearby analog connect
Answer:
Crosstalk
Explanation:
The answer is Crosstalk as this phenomenon is most commonly associated with analog phone call.
Now, crosstalk is defined as a disturbance caused by the electric or magnetic fields of one telecommunication signal which affects a signal in an adjacent circuit. In a telephone circuit, crosstalk could result in hearing part of a voice conversation from another circuit. Hence, the phenomenon that causes crosstalk is called electromagnetic interference (EMI). This may occur in microcircuits within computers and audio equipments including within network circuits. This term is also usually applied to optical signals that interfere with each other.
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
A dipole moment is placed in a uniform electric field oriented along an unknown direction. The maximum torque applied to the dipole is equal to 0.1 N.m. When the dipole reaches equilibrium its potential energy is equal to -0.2 J. What was the initial angle between the direction of the dipole moment and the direction of the electric field?
Answer:
θ = 180
Explanation:
When an electric dipole is placed in an electric field, there is a torque due to the electric force
τ = p x E
by rotating the dipole there is a change in potential energy
ΔU = ∫ τ dθ
ΔU = p E (cos θ₂ - cos θ₁)
when the dipole starts from an angle to the equilibrium position for θ = 0
ΔU = pE (cos θ - cos 0)
cos θ = 1 + DU / pE)
let's apply this expression to our case, the change in potential energy is ΔU = -0.2J
let's calculate
cos θ = 1 -0.2 / 0.1
cos θ = -1
θ = 180
A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that numerically equal to the instantaneous velocity. The mass is initially released from a point 1 ft above the equilibrium position with a downward velocity of 14 feet/second. Determine the time at which the mass passes through the equilibrium position
Answer:
the time at which it passes through the equilibrum position is:
t = 0.1 second
Explanation:
given
w= 4pounds
k(spring constant) = 2lb/ft
g(gravitational constant) = 10m/s² = 32ft/s²
β(initial point above equilibrum) = 1
velocity = 14ft/s
attached is an image showing the calculations, because some of the parameters aren't convenient to type.
The time at which the mass passes will be "0.1 s".
EquilibriumAccording to the question,
Mass weighing, w = 4 pounds
Spring constant, k = 2 lb/ft
Gravitational constant, g = 10 m/s² = 32 ft/s²
Point above equilibrium, β = 1
Velocity = 14 ft/s
By using equation of motion,
→ x(t) = (-1 + gt)
By substituting the values,
0 = (-1 + 10t)
-1 + 10t = 0
By adding "1" both sides, we get
-1 + 10t + 1 = 1
10t = 1
t = [tex]\frac{1}{10}[/tex]
= 0.1 s
Thus the above answer is right.
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When a boat is placed in liquid, two forces act on the boat. Gravity pulls the boat down with a force equal to the weight of the boat. Weight is measured in newtons (N). To calculate the weight of a boat, multiply its mass in grams by 0.00982.
As the boat sinks into the liquid, the liquid pushes back. The force of the liquid pushing up on the boat is called the buoyant force.
Required:
How do gravity and the buoyant force affect a boat?
Answer:
the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
Explanation:
This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law
B- W = 0
B = W
the thrust force is equal to the weight of the liquid that is dislodged
B = ρ g V
we substitute
ρ g V = m g
V = m /ρ_fluid 1
we can write the mass of the pot as a function of its density
ρ_body = m / V_body
m = ρ_body V_body
V_fluid / V_body = ρ_body / ρ _fluid 2
Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
The effect appears the pot as if it had a lower apparent weight
The cable of a crane is lifting a 950 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/sin a distance of 5.5m
a) How much work is done by gravity?
b) How much work is done by tension?
Explanation:
a) How much work is done by gravity?
w = f x dw = 950 x 10 x 5.5 = 52250jb) How much work is done by tension?
v²=u²+2as0.75²=0.25²+2a x5.50.56=0.06+2a x5.52a x5.5 = 0.56 - 0.062a x 5.5 =0.511a=0.5 a = 0.5/11 = 0.05m/s²w = f x d
w = 950 x 0.05 x 5.5 = 261.25j
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an overall angular magnification of 300
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is [tex]f_e = 0.027 \ m[/tex]
Explanation:
From the question we are told that
The focal length is [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]
This negative sign shows the the microscope is diverging light
The angular magnification is [tex]m = 300[/tex]
The distance between the objective and the eyepieces lenses is [tex]Z = 19 \ cm = 0.19 \ m[/tex]
Generally the magnification is mathematically represented as
[tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]
Where [tex]f_e[/tex] is the eyepiece focal length of the microscope
Now making [tex]f_e[/tex] the subject of the formula
[tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]
substituting values
[tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]
[tex]f_e = 0.027 \ m[/tex]
Victoria’s ambition is to climb Mount Everest, also known in Nepal as Sagarmatha and in Tibet as Chomolungma. She thinks that the effect of the changing gravitational force will mean that she will find the climb easier as she climbs higher. The Earth’s radius is approximately 6400 km. The height of mount Everest is approximately 8.8 km. a. Explain how the gravitational force between two objects changes with the distance between them. b. With reference to the data above and to what you know about gravitational force field, evaluate Victoria’s prediction. c. Calculate the height above the Earth’s surface at which Victoria’s weight would be half its value on the surface
Answer:
When these objects moves away or come close together the change occurs in force of gravity.
Explanation:
When these objects moves away or come close together the gravitational force between two objects changes with the distance between them . At the center of the earth gravitational force is the highest while moving away from the center, the force of gravity decreases. Yes, Victoria’s prediction was right about gravitational force. When we go to a specific height, the force of gravity decreases because we know that gravitational force decreases with increasing height. Victoria’s weight would be half its value on the surface when she is present at the altitude of 2642 km because at that altitude force of gravity is also half.
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2 cm^2. A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is 9.20 N⋅m^2/C, what area of the sheet is enclosed by the Gaussian surface?
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.
Charge and Charge DensityA certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.
Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.
[tex]\sigma = \dfrac {Q}{A}[/tex]
[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]
[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]
Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].
Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.
[tex]Q' = \sigma A'[/tex]
Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.
For the closed Gaussian Surface, Flux is given below.
[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]
Hence the charge can be written as,
[tex]Q' = \phi\epsilon_\circ[/tex]
So the charge can be given as below.
[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]
Then the surface area of the Gaussian surface is given below.
[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]
Substituting the values in the above equation,
[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]
[tex]A' =0.006076\;\rm m^2[/tex]
[tex]A' = 60.76 \;\rm cm^2[/tex]
Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.
To know more about the charge and charge density, follow the link given below.
https://brainly.com/question/8532098.
A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that passes through the center and lies in the plane of the loop. When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 N ⋅ m. What is the magnitude of the uniform magnetic field exerting this torque on the loop?
Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :
[tex]\tau=NIAB\sin\theta[/tex]
B is magnetic field
[tex]B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T[/tex]
So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.
Which of the following statements is valid:_______.A) If the potential throughtout a particular region is constant, the field throughout the region must be zero.B) If the field throughout a particular region is constant, the potential throughout that region must be zero.C) If the field at a particular point is zero, the potential at that point must be zero.D) If the potential at a particular point is zero, the field at that point must be zero.
Answer:
Option A is the only valid statement.
Explanation:
A)The electric field intensity is defined by the relationship:
E= -ΔV/Δr.
Now, according to the relationship above, the electric field would be the negative gradient of electric potential. Now, if the electric potential is constant throughout the given region of space, then the change in electric potential would be ΔV=0.
Thu,E= 0.
So the answer is that, E will be zero in this case.
So, the statement is valid.
B) Statement not valid because the field is the gradient of the potential. Hence, the field would be zero in any region where the potential is constant. However, constant does not necessarily mean a value of zero. With that being said, we can always change the definition of the potential function by adding a constant, to thus make it zero there. But then the potential will no longer be zero at infinity or in any different “flat” regions.
C) Statement not valid because, for the fact that electric field is zero at a particular point, it doesn't necessarily
imply that the electric potential is zero at that point. A good example would be the case of two identical charges which are separated by some distance. At the midpoint between the charges, the
electric field due to the charges would be zero. However, the electric potential due to the charges at that same point would not be zero. Thus, the potential will either have two positive contributions, if the charges are positive, or two negative contributions, if the charges are negative.
(D) Statement is not valid because, for the fact that electric potential is zero at a particular point, it does not necessarily imply that the electric field is zero at that point. A good example would be the case of a dipole, which
has two charges of the same magnitude, but opposite sign, and are separated by some distance. At
the midpoint between the charges, the electric potential due to the charges would be zero, but the electric field due to the charges at that same point would not be zero.
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical nerve cell, 9.0 pC of charge flows in a time of 0.50 ms. You may want to review (Page 706) . Part A What is the average current
Answer:
I = 18 x 10⁻⁹ A = 18 nA
Explanation:
The current is defined as the flow of charge per unit time. Therefore,
I = q/t
where,
I = Average Current passing through nerve cell
q = Total flow of charges through nerve cell
t = time period of flow of charges
Here, in our case:
I = ?
q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C
t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s
Therefore,
I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)
I = 18 x 10⁻⁹ A = 18 nA
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.