Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and 300 K, respectively. The heat transfer from the air during the isothermal compression is 80 kJ. At the end of the isothermal compression, the volume is 0.2 m3. Determine the volume at the beginning of the isothermal compression, in m3. Assume the ideal gas model for air and neglect kinetic and potential energy effects.
Answer:
Explanation:
Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .
So the given process of isothermal compression must have been done at the temperature of 300K , keeping the temperature constant .
Work done on gas at isothermal compression is equal to heat transfer .
work done on gas = 80 x 10³ J
work done on gas = n RT ln v₁ / v₂
n is number of moles v₁ and v₂ are initial and final volume
molecular weight of gas = 28.97 g
1.5 kg = 1500 / 28.97 moles
= 51.77 moles
work done on gas = n RT ln v₁ / v₂
Putting the values in the equation above
80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2
ln v₁ / .2 = .62
v₁ / .2 = 1.8589
v₁ = 0.37 m³
A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 15 m/s. The ball rebounds at 40 m/s.
A) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
B) If the tennis ball and racket are in contact for 7.00, what is the average force that the racket exerts on the ball?
_________N
The velocity and force are required.
The speed of the racket is 8.7 m/s
The required force is 471.43 N.
[tex]m_1[/tex] = Mass of racket = 1000 g
[tex]m_2[/tex] = Mass of ball = 60 g
[tex]u_1[/tex] = Initial velocity of racket = 12 m/s
[tex]u_2[/tex] = Initial velocity of ball = -15 m/s
[tex]v_1[/tex] = Final velocity of racket
[tex]v_2[/tex] = Final velocity of ball = 40 m/s
[tex]\Delta t[/tex] = Time = 7 ms
The equation of the momentum will be
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow v_1=\dfrac{m_1u_1+m_2u_2-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{1\times 12+0.06\times (-15)-0.06\times 40}{1}\\\Rightarrow v_1=8.7\ \text{m/s}[/tex]
Force is given by
[tex]F=m_2\dfrac{v_2-u_2}{\Delta t}\\\Rightarrow F=0.06\times \dfrac{40-(-15)}{7\times 10^{-3}}\\\Rightarrow F=471.43\ \text{N}[/tex]
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show that energy dissipated due to motion of a conductor in the magnetic field is due to mechanical energy.
Answer:
P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.
Explanation:
The force, F on a conductor of length, L in a magnetic field of magnetic field strength, B and current, I flowing through it is given by
F = BIL.
Now, if the conductor has a velocity, v, the energy dissipated by this force is P
P = Fv = BIL × v = BILv.
Now, we know that the induced e.m.f due to the motion of the conductor is given by ε = BLv.
From P above, P = BILv = I(BLv)
substituting ε = BLv into P, we have
P = Iε
Thus, P is the electrical energy dissipated due to the motion of the conductor.
Now since P = BILv = Iε, it shows that the mechanical energy due to the force on the conductor equals the electrical energy dissipated due to the motion of the conductor in the magnetic field and so, energy is conserved.
A coil of 160 turns and area 0.20 m2 is placed with its axis parallel to a magnetic field of initial magnitude 0.40 T. The magnetic field changes uniformly from 0.40 T in the +x direction to 0.40 T in the -x direction in 2.0 s. If the resistance of the coil is 16 Ω, at what rate is power generated in the coil?
Answer:
The rate at which power is generated in the coil is 10.24 Watts
Explanation:
Given;
number of turns of the coil, N = 160
area of the coil, A = 0.2 m²
magnitude of the magnetic field, B = 0.4 T
time for field change = 2 s
resistance of the coil, R = 16 Ω
The induced emf in the coil is calculated as;
emf = dΦ/dt
where;
Φ is magnetic flux = BA
emf = N (BA/dt)
emf = 160 (0.4T x 0.2 m²)/dt
emf = 12.8 V/s
The rate power is generated in the coil is calculated as;
P = V²/ R
P = (12.8²) / 16
P = 10.24 Watts
Therefore, the rate at which power is generated in the coil is 10.24 Watts
A container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37, what is the index of refraction of the second fluid
Answer:
The refractive index of fluid 2 is 1.78
Explanation:
Refractive index , n = real depth/apparent depth
For the first fluid, n = 1.37 and apparent depth = 9.00 cm.
The real depth of the container is thus
real depth = n × apparent depth = 1.37 × 9.00 cm = 12.33 cm
To find the refractive index of fluid index of fluid 2, we use the relation
Refractive index , n = real depth/apparent depth.
Now,the real depth = 12.33 cm and the apparent depth = 6.86 cm.
So, n = 12.33 cm/6.86 cm = 1.78
So the refractive index of fluid 2 is 1.78
Since the same container is used, real depth of fluid 1 is equal to the real depth of fluid 2. The index of refraction of the second fluid is 1.8
Given that a container is filled with fluid 1, and the apparent depth of the fluid is 9.00 cm. The container is next filled with fluid 2, and the apparent depth of this fluid is 6.86 cm. If the index of refraction of the first fluid is 1.37,
Then,
Index of refraction = [tex]\frac{Real depth}{Apparent depth}[/tex]
Real depth = Index of refraction x apparent depth
Since the same container is used, we can make an assumption that;
real depth of fluid 1 = real depth of fluid 2
That is,
1.37 x 9 = n x 6.86
Where n = Index of refraction for the second fluid.
make n the subject of formula
n = 12.33 / 6.86
n = 1.79
Therefore, the index of refraction of the second fluid is 1.8 approximately.
Learn more about refraction here: https://brainly.com/question/10729741
A bug is sitting on the edge of a rotating disk. At what angular velocity will the bug slide off the disk if its radius is 0.241 m, the coefficient of static friction between the bug and disk is 0.321, and the coefficient of kinetic friction is 0.102
Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
what is rotation? a. the orbit of a satelite around a central body b. the motion of two objects around each other c the spinning of an object on its axis d the motion of an object around a central body
Answer:
C. The spinning of an object on its axis.
Explanation:
The definition of rotation answers this question.
A good way to remember is that when you rotate, you turn, and when you revolve, you move around.
A cylindrical rod 120 mm long and having a diameter of 15 mm is to be deformed using a tensile load of 35,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.2x10-2 mm. Of the materials listed below, which are possible candidates? Justify your choice(s).
Material Modulus of Elasticity (GPa) Yield Strength (MPa) Poisson's ratio
Titanium alloy 70 250 0.33
Steel alloy 105 850 0.36
Magnesium alloy 205 550 0.27
Aluminium alloy 45 170 0.20
Answer:
The only suitable candidate is steel.
Explanation:
We are given;
Initial length;L_o = 120 mm = 0.12 m
Diameter;d_o = 15 mm = 0.015 m
Radius;r = 0.015/2 m = 0.0075 m
Applied force;F = 35000 N
Allowable Diameter reduction;Δd = 1.2 x 10^(-2) mm = 0.012 mm
Formula for Poisson’s ratio is;
v = ε_x/ε_z
Where;
ε_x is lateral strain = = Δd/d_o
ε_z is axial strain = σ/E
Thus;
v = (Δd/d_o)/(σ/E)
Making Δd the subject, we have;
Δd = (vσd_o)/E
Where σ is force per unit area
σ = F/A = 35000/πr² = 35000/(π0.0075)² = 198 × 10^(6) Pa
Now, let's find Δd for each of the metals given;
For Titanium alloy, E = 70 GPa = 70 × 10^(9) Pa and Poisson’s ratio (v) = 0.33
Thus;
Δd_ti = (0.33 × 198 × 10^(6) × 0.015)/70 × 10^(9) = 0.000014m = 0.014 mm
Similarly, for steel;
Δd_st = (0.36 × 198 × 10^(6) × 0.015)/(105 × 10^(9)) = 0.0000102 m = 0.0102 mm
For magnesium;
Δd_mn = (0.27 × 198 × 10^(6) × 0.015)/(205 × 10^(9)) = 0.0000039 m = 0.0039 mm
For aluminum;
Δd_al = (0.2 × 198 × 10^(6) × 0.015)/(45 × 10^(9)) = 0.0000132 m = 0.0132 mm
Since they must not experience a diameter reduction of more than 0.012 mm, then the only alloy that has a reduction diameter less than 0.012 is steel which has 0.0102.
Thus,the only possible candidate is stee.
In Young's double slit experiment if the maximum intensity of light is Imax, then the intensity at path difference λ/2 will be
A .Imax
B .2Imax
C .4Imax
D. None
Answer:
Explanation:
When the path difference is half the wave length or λ /2 , destructive interference takes place which results in reduced or zero intensity in case equal intensity waves interfere as in Young's double slit experiment
Hence dark fringe is formed at that place where intensity is zero .
Hence the right option is D
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
A spring balance is attached with string to the piece of aluminum in the preceding problem. What reading will the balance register when the metal is submerged
A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.
[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.
Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]
And final speed is now calculated after clearing it:
[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]
[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have
Answer:
50000 turns
Explanation:
Vp / Vs = Np / Ns
240 / 12000 = 1000 / Ns
Ns = 50000 turns
The AC voltage source supplies an rms voltage of 146 V at frequency f. The circuit has R = 110 Ω, XL = 210 Ω, and XC = 110 Ω. At the instant the voltage across the generator is at its maximum value, what is the magnitude of the current in the circuit?
Answer:
1.03A
Explanation:
For computing the magnitude of the current in the circuit we need to do the following calculations
LCR circuit impedance
[tex]Z = \sqrt{R^2 + (X_L - X_c)^2} \\\\ = \sqrt{110^2 + (210 - 110)^2}[/tex]
= 148.7Ω
Now the phase angle is
[tex]\phi = tan^{-1} (\frac{X_L - X_C}{R}) \\\\ = tan^{-1} (\frac{210 - 110}{110})\\\\ = 42.3^{\circ}[/tex]
Now the rms current flowing in the circuit is
[tex]I_{rms} = \frac{V_{rms}}{Z} \\\\ = \frac{146}{148.7}[/tex]
= 0.98 A
The current flowing in the circuit is
[tex]I = I_{rms}\sqrt{2} \\\\ = (0.98) (1.414)[/tex]
= 1.39 A
And, finally, the current across the generator is
[tex]I'= I cos \phi[/tex]
[tex]= (1.39) cos 42.3^{\circ}[/tex]
= 1.03A
Hence, the magnitude of the circuit current is 1.03A
2. In the ice cream making process, after the pasteurization of the base mixture, the syrup should be cooled to 4 °C to avoid the proliferation of pathogenic microorganisms. A new thermometer was attached to the tank; however, it marked a temperature in another unit: Rankine. What should be the value indicated on the thermometer for the process to be carried out under the same conditions?
3. During the class in the laboratory, the manometer coupled to the analysis equipment indicates a vacuum of 638 mmHg. What should be the absolute pressure in kPa and psi, knowing that the local barometric pressure is 101.3 kPa?
Answer:
2. 500 R
3. 16.3 kPa, 2.36 psi
Explanation:
2. Convert Celsius to Fahrenheit.
1.8 (4°C) + 32 = 39.2°F
Convert Fahrenheit to Rankine
39.2°F + 459.67 = 498.87 R
Rounding to one significant figure, the temperature is 500 R.
3. Absolute pressure = gauge pressure + atmospheric pressure
P = Pg + Pa
First, convert mmHg to kPa (remember that a vacuum is negative gauge pressure).
-638 mmHg × (101.3 kPa / 760 mmHg) = -85.0 kPa
So the absolute pressure is:
P = -85.0 kPa + 101.3 kPa
P = 16.3 kPa
Converting to psi:
P = 16.3 kPa × (14.7 psi / 101.3 kPa)
P = 2.36 psi
A man has vocal cords of length 22 mm, with a mass per length of 0.0042 kg/m. What tension is required in the vocal cords in order to produce a tone of middle C (261.62 Hz)?
Answer:
Tension, T = 0.556 N
Explanation:
It is given that,
Length of vocal cords, l = 22 mm = 0.022 m
Mass per unit length, [tex]\mu=0.0042\ kg/m[/tex]
We need to find the tension is required in the vocal cords in order to produce a tone of middle C of frequency 261.62 Hz. The frequency in terms if tension is given by :
[tex]f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/tex]
T = tension in the vocal cords
[tex]f^2=\dfrac{1}{4l^2}\times \dfrac{T}{\mu}\\\\T=4l^2\mu f^2\\\\T=4\times (0.022)^2\times 0.0042 \times (261.62 )^2\\\\T=0.556\ N[/tex]
So, the tension in the vocal cords is 0.556 N.
A customs inspector was suspecting that some of the 12 plastic spheres, which were shipped out of the country, had something in them. Each sphere weighted the same and had hard walls everywhere. Inspector thought that it was possible to hide something inside each sphere. He was correct, and was able to use a simple experiment in determining which sphere had diamonds inside. How did he do it?
Answer:
use a hammer to hit it
Explanation:
if u hit it u will be able to hear the shattered noise
The moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is: . If the axis of rotation passes through the center of the rod, then the moment of inertia is: . Give a physical explanation for this difference in terms of the way the mass of the rod is distributed with respect to the axis in the two cases.
Answer:
Explanation:
he moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is: m L²/ 3 where m is mass and L is length of rod
If the axis of rotation passes through the center of the rod, then the moment of inertia is: m L² / 12
So for the former case , moment of inertia is higher that that in the later case .
In the former case , the axis is at one extreme end . Hence range of distance of any point on the rod from axis is from zero to L .
In the second case , as axis passes through middle point , this range of distance of any point on the rod from axis is from zero to L / 2 .
Since range of distance from axis is less , moment of inertia too will be less because
Moment of inertia = Σ m r² where r is distance of mass m from axis .
gravity can be described as..?
A. an magnetic force found in nature
B.the force that moves electrical charges
C.the force that repels object with like chargers
D.the force of attraction between two objects
Answer:
D
Explanation:
Gravity is the force of attraction between two objects.
Each object creates a gravitational field in wich every other object is affected by it.
If the magnetic field steadily decreases from BBB to zero during a time interval ttt, what is the magnitude III of the induced current
Answer:
Using ohms law
The current is found from Ohm's Law.
I = V /R = E /R = Bxy /Rt.
Determine the slit spacing d. Explain which measurement you made, show your calculation and your result for the slit spacing. There are several measurements you can make.
Answer:
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
Explanation:
To determine the gap spacing we must use the constructive interference equation
d sin θ = m λ
as the angles are small
tan θ = sin θ / cos θ
tan θ = sin θ
and the definition of tangent is
tan θ = y / L
Thus
sin θ = y / L
when replacing
d y / L = m λ
d = m λ L / y
with this equation we can know what parameter should be measured.
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
A spring is attached to the ceiling and pulled 11 cm down from equilibrium and released. The amplitude decreases by 19% each second. The spring oscillates 10 times each second. Find an equation for the distance, D the end of the spring is below equilibrium in terms of seconds, t.
Answer:
D(t)= 11x2.09t cos( 20pi,t)
Explanation:
Pls attached file
A 22.0 cm diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its plane is parallel to the field direction in 0.20 s. What is the average induced emf in the loop?
Answer:
The average induced emf in the loop is 0.3 V
Explanation:
Given:
d = 22 cm
Magnetic field is B = 1.5 T
Change in time Δt = 0.20 sec.
Radius of loop = r = d/2 = 11 x [tex]10^{-2}[/tex] m
According to the faraday's law, Induced emf is given by e = - (ΔФ / Δt)
where Ф = magnetic flux
Ф = BAcos0 (in this occasion, Ф=0)
where area A = pi * r²
note that we have to neglect negative sign due to its lenz law...so
B * pi * r
e = ----------------
Δt
1.5 ( 3.1416) 11 x [tex]10^{-2}[/tex] )²
e = ------------------------------------
0.20
e = 0.3 V
Therefore, the average induced emf in the loop is 0.3 V
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?
Answer:
The wait time is [tex]t_w = 3.4723 \ s[/tex]
Explanation:
From the question we are told that
The distance of the hot air balloon above the ground is [tex]z = 50 \ m[/tex]
The distance of the hot air balloon from the target is [tex]k = 100 \ m[/tex]
The speed of the wind is [tex]v = 15 \ m/s[/tex]
Generally the time it will take the balloon to hit the ground is
[tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]
where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]
[tex]t = 3.194 \ s[/tex]
Now at the velocity the distance it will travel before it hit the ground is mathematically represented as
[tex]d = v * t[/tex]
substituting values
[tex]d = 15 * 3.194[/tex]
[tex]d = 47.916 \ m[/tex]
Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as
[tex]b = k - d[/tex]
substituting values
[tex]b =100 -47.916[/tex]
[tex]b = 52.084 \ m[/tex]
Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as
[tex]t_w = \frac{b}{v}[/tex]
substituting values
[tex]t_w = \frac{52.084}{15}[/tex]
[tex]t_w = 3.4723 \ s[/tex]
A 150 g ball rolls at 10 cm/s rightward over a frictionless surface towards a spring 75 cm away, with spring constant 200 N/m. How far does is the spring compressed when the ball is brought to rest
Answer:
x = 0.27 cm
Explanation:
given data
mass = 150 g
velocity = 10 cm/s = 0.1 m/s
spring constnat = 200 N/m
solution
as we know that here ball is moving with constant speed
so
0.5 × m × v² = 0.5 × k × x² .......................1
here x is compression in spring
so put here value and we get
0.5 × 150 × (0.1)² = 0.5 × 200 × x²
solve it we get
x = 0.27 cm
A charge of 0.80 nC is placed at the center of a cube that measures 4.0 m along each edge. What is the electric flux through one face of the cube
Answer:
The magnetic flux is [tex]\phi = 15 \ Nm^2 /C[/tex]
Explanation:
From the question we are told that
The value of the charge is [tex]q = 0.80 \ nC = 0.80 *10^{-9} \ C[/tex]
The length of each side of the cube is [tex]d = 4.0 \ m[/tex]
Generally the magnetic flux through a closed surface is mathematically represented as
[tex]\phi = \frac{q}{\epsilon_o} * D[/tex]
Where D is the area enclosing the charge
Now a cube is made up of six faces but in this question we are considering only one face which is mathematically represented as
[tex]D = \frac{1}{6}[/tex]
So the electric flux through one face of the cube is mathematically represented as
[tex]\phi = \frac{q}{6 * \epsilon _o }[/tex]
where [tex]\epsilon _o[/tex] is the permitivity of free space with value
[tex]\epsilon_o = 8.85 *10^{-12} F/m[/tex]
substituting value
[tex]\phi = \frac{0.80 *10^{-9}}{6 * 8.85 *10^{-12} }[/tex]
[tex]\phi = 15 \ Nm^2 /C[/tex]
To work on your car at night, you use an extension cord to connect your work light to a power outlet near the door. How would the illumination provided by the light be affected by the length of the extension cord
Answer:
The longer the cord, the lower the illumination
Explanation:
The illumination provided by the light bulb will be reduced as the length of the extension cord increases. This is because the resistance provided by the wire increases with its length.
Long wires have more electrical resistance than shorter ones.
Let us consider this formula:
Resistance =[tex]\frac{\rho L}{A}[/tex]
From this formula, we can see that as the length increases, the resistance to current flow offered by the wire increases also provided the resistivity and cross-sectional area of the wire remain constant. As a result of this, the illumination will drop.
Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as originating from a point on the near side of the mirror.
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved mirror than the centre of curvature.
During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutual force between them
Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
[tex]F = \frac{k|q_1||q_2|}{r^2}[/tex]
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
[tex]F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N[/tex]
Therefore, the mutual force between the two point charges is 319.64 N
A millionairess was told in 1992 that she had exactly 15 years to live. However, if she immediately takes off, travels away from the Earth at 0.8 c and then returns at the same speed, the last New Year's Day the doctors expect her to celebrate is:
Answer:
The expected year is 2017.
Explanation:
Total years that the millionaire to live = 15 years
Travel away from the earth at = 0.8 c
This is a time dilation problem so if she travels at 0.8 c then her time will pass at slower. Below is the following calculation:
[tex]T = \frac{T_o}{ \sqrt{1-\frac{V^2}{c^2}}} \\T = \frac{15}{ \sqrt{1-\frac{0.8^2}{c^2}}} \\T = 25 years[/tex]
Thus the doctors are expecting to celebrate in the year, 1992 + 25 = 2017