Answer:
The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]
Explanation:
From the question we are told
The diameter of the loop is [tex]d = 6.7 cm = 0.067 \ m[/tex]
The magnetic field is [tex]B = 1.27 \ T[/tex]
The time taken is [tex]dt = 0.16 \ s[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]
Where N = 1 given that it is only a circular loop
[tex]\Delta \phi = \Delta B * A[/tex]
Where [tex]\Delta B = B_f - B_i[/tex]
where [tex]B_i[/tex] is 1.27 T given that the loop of wire was initially in the magnetic field
and [tex]B_f[/tex] is 0 T given that the loop was removed from the magnetic field
Now the area of the of the loop is evaluated as
[tex]A = \pi r^2[/tex]
Where r is the radius which is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.067}{2}[/tex]
[tex]r = 0.0335 \ m[/tex]
So
[tex]A = 3.142 * (0.0335)^2[/tex]
[tex]A = 0.00353 \ m^2[/tex]
So
[tex]\Delta \phi = (0- 127)* (0.00353)[/tex]
[tex]\Delta \phi = -0.00448 Weber[/tex]
=> [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]
=> [tex]\epsilon = 0.0280 \ V[/tex]
Explain why water, with its high specific heat capacity, is utilized for heating systems such as hot-water radiators.
Answer:
Answer in explanation
Explanation:
Water is mainly used as coolant in heating systems like hot-water radiators. The main function of water in such systems, is to absorb as much heat as possible, in order to decrease the temperature of the system and as a result cool it.
The specific heat capacity is the measure of heat energy that is required to raise the temperature of unit mass of a substance through 1 °C. In other words, specific heat capacity quantifies the amount of heat that can be stored by a unit mass of a substance having a degree rise in temperature.
Thus, the more specific heat a substance has, the more heat it can absorb from the hot system. Hence, the specific heat capacity of a coolant must be high.
This is the reason why water, with its high specific heat capacity, is utilized for heating systems, such as radiators.
B. CO
A wave has frequency of 2 Hz and a wave length of 30 cm. the velocity of the wave is
A. 60.0 ms
B. 6.0 ms
D. 0.6 ms
Answer:
0.6 m/s
Explanation:
2Hz = 2^-1 = 2 /s
30cm = .3m
Velocity is in the units m/s, so multiplying wavelength in meters by the frequency will give you the velocity.
(.3m)*(2 /s) = 0.6 m/s
A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0215 W/m2W/m2, and the wavelength of the wave is 6.90 mm.What is the maximum emf induced in the loop?
Express your answer with the appropriate units.
Answer:
The induced emf is [tex]\epsilon = 0.1041 \ V[/tex]
Explanation:
From the question we are told that
The radius of the circular loop is [tex]r = 9.50 \ cm = 0.095 \ m[/tex]
The intensity of the wave is [tex]I = 0.0215 \ W/m^2[/tex]
The wavelength is [tex]\lambda = 6.90\ m[/tex]
Generally the intensity is mathematically represented as
[tex]I = \frac{ c * B^2 }{ 2 * \mu_o }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]
B is the magnetic field which can be mathematically represented from the equation as
[tex]B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }[/tex]
substituting values
[tex]B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }[/tex]
[tex]B = 1.342 *10^{-8} \ T[/tex]
The area is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * (0.095)^2[/tex]
[tex]A = 0.0284[/tex]
The angular velocity is mathematically represented as
[tex]w = 2 * \pi * \frac{c}{\lambda }[/tex]
substituting values
[tex]w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }[/tex]
[tex]w = 2.732 *10^{8} rad \ s^{-1}[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = N * B * A * w * sin (wt )[/tex]
At maximum induced emf [tex]sin (wt) = 1[/tex]
So
[tex]\epsilon = N * B * A * w[/tex]
substituting values
[tex]\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}[/tex]
[tex]\epsilon = 0.1041 \ V[/tex]
An array of solar panels produces 9.35 A of direct current at a potential difference of 195 V. The current flows into an inverter that produces a 60 Hz alternating current with Vmax = 166V and Imax = 19.5A.
A) What rms power is produced by the inverter?
B) Use the rms values to find the power efficiency Pout/Pin of the inverter.
Answer:
(A). 1620 watt.
(B).0.8885.
Explanation:
So, we are given the following data or parameters or information which is going to assist or help us in solving this particular Question or problem. So, we have;
Current = 9.35A, direct current at a potential difference of 195 V, frequency of the inverter = 60 Hz alternating current, alternating current with Vmax = 166V and Imax = 19.5A.
(A). The rms power is produced by the inverter = (19.5 /2 ) × 166 = 1620 watt(approximately).
(B). the rms values to find the power efficiency Pout/Pin of the inverter.
P(in) = 195 × 9.35 = 1823.3 watt.
Thus, the rms values to find the power efficiency Pout/Pin of the inverter = 1620/1823.3 = 0.88852324146441793 = 0.8885.
Consider two isolated spherical conductors each having net charge Q. The spheres have radii a and b, where b > a.
Which sphere has the higher potential?
1. the sphere of radius a
2. the sphere of radius b
3. They have the same potential
Answer:
1. the sphere of the radius a
Explanation:
Because the charge distribution for each case is spherically symmetric, we can choose a spher- ical Gaussian surface of radius r , concentric with the sphere in question.
So E = k (Q /r 2) (for r ≥ R ) , where R is the radius of the sphere being considered, either a or b .
With the choice of potential at r = ∞ being zero, the electric potential at any distance r from the center of the sphere can be expressed as V = - integraldisplay r ∞ E dr = k /Q r
(for r ≥ R ) .
On the spheres of radii a and b , we have V a = k (Q/ a)and V b = k (Q/ b), respectively.
So Since b > a , the sphere of radius a will have the higher potential.
Also recall Because E = 0 inside a conductor, the potential
"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"
Answer:
The film thickness is 4.32 * 10^-6 m
Explanation:
Here in this question, we are interested in calculating the thickness of the film.
Mathematically;
The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;
ΔN = (2L/λ) (n-1)
where λ is the wavelength of the light used
Let’s make L the subject of the formula
(λ * ΔN)/2(n-1) = L
From the question ΔN = 8 , λ = 540 nm, n = 1.5
Plugging these values, we have
L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m
A person starts at position zero, walks to position 8, then walks to position 5. Which answer correctly identifies the person's distance traveled? *
Answer:
Distance = 13 units
Explanation:
The overall path covered by an object during its journey is called distance covered.
In this problem, a person starts at position zero, walks to position 8, then walks to position 5.
We need to find the person's distance traveled. It can be calculated simply by adding all the positions i.e.
Distance = 0+8+5
Distance = 13
Hence, the distance covered by the person is 13 units.
A football is kicked with a velocity of 18 m/s at an angle of 20°. What is the
ball's acceleration in the horizontal direction as it flies through the air?
Explanation:
It is given that,
The velocity of football is 18 m/s
It is projected at an angle of 20 degrees
We need to find the ball's acceleration in the horizontal direction as it flies through the air.
When it is projected with some velocity, it has two rectangular components i.e. horizontal and vertical.
In vertical direction, it will move under the action of gravity. There is no change in velocity in horizontal direction. So, ball's acceleration in the horizontal direction is equal to 0.
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective? Why?
(b) How large a sail is necessary to propel a 10000kg
spacecraft against the gravitational force of the sun? Express your result in square kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun.
The gravitational constant is G=6.67×10−11m3⋅s−2⋅kg−1.
The mass of the sun is Ms=1.99×1030kg.
Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
Large capacitors can hold a potentially dangerous charge long after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 μF capacitor from a camera flash unit retains a voltage of 140 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Answer:
93.3x10^-3s
Explanation:
If
Resistance = 1.8 kΩ
Current = 50 mA
Capacitor = 120 μF
Voltage = 140 V
to calculate the discharge current
Applying the formula of discharge current
io=vo/R
io= 140/ 1.8x 10³
= 0.078A
to calculate the time
Applying the formula of current
io= vo/R e-t/RC
50= 140/1800e-t/RC
0.649= e-t/RC
-t/RC= ln( 0.649)
t = 0.432x 120x10^-6x 1800
t=93.3 x 10^-3seconds
In France, the wall sockets provide an AC voltage with Vrms = 230 V. You want to use an appliance designed to operate in the United States (Vrms = 120 V) and decide to build a transformer to convert the power line voltage in France to the value required by your appliance.
(a) Should you use a "step-down" transformer (to make Vrms smaller) or a "step-up" transformer (which makes Vrms larger)?
a "step-up" transformer
a "step-down" transformer
(b) If the input coil of your transformer has 2760 turns, how many turns should the output coil have?
_____ turns
Answer:
a)step-down" transformer
b) 1440 turns
Explanation:
There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.
In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.
If
Number of turns in primary coil Np= 2760
Number of turns in secondary coil Ns= unknown
Voltage in primary coil Vp= 230 V
Voltage in secondary coil Vs= 120 V
Ns/Np= Vs/VP
NsVp= NpVs
Ns= NpVs/VP = 2760 × 120/230
Ns= 1440 turns
Describe at least two unique characteristics of the new space telescope.
Answer: Astro 1 will have a 10x larger field of view than the Hubble.
Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.
There are two cells, one with OER as 2.5 and other as 7. Which cell is more sensitive to radiation?
1)The cell with OER 2.5
2)The Cell with OER 7
3)Both the cells
4)Insufficient data
Answer:
2)The Cell with OER 7
Explanation:
OER is the acronym for Oxygen Enhancement Ratio. It is the measure of the enhancement of the effect of ionizing radiation due to the presence of oxygen. The ionization effect can be detrimental or therapeutic (use in cancer treatment). OER is the ratio of radiation dose during the lack of oxygen (hypoxia), to the dosage in the presence of oxygen (air is used as a reference). From the definition, one can see that the higher the OER the higher the sensitivity of the cell.
A mother and her young child want to play on a seesaw at a playground. The child sits on the end of one side of the seesaw. Where should the mother sit to balance the seesaw?(1 point) at the opposite side of the seesaw on the end at the opposite side of the seesaw towards the middle on the same side of the seesaw towards the middle on the same end as her child
Answer:middle
Explanation:
Because it will make the seasaw balanced
An oil film (n = 1.48) of thickness 290 nm floating on water is illuminated with white light at normal incidence. What is the wavelength of the dominant color in the reflected light? A. Blue (470 nm) B. Green (541 nm) C. Violet (404 nm) D. Yellow (572 nm)
Answer:
The correct option is D
Explanation:
From the question we are told that
The refractive index of oil film is [tex]k = 1.48[/tex]
The thickness is [tex]t = 290 \ nm = 290*10^{-9} \ m[/tex]
Generally for constructive interference
[tex]2t = [m + \frac{1}{2} ]* \frac{\lambda}{k}[/tex]
For reflection of a bright fringe m = 1
=> [tex]2 * (290*10^{-9}) = [1 + \frac{1}{2} ]* \frac{\lambda}{1.48}[/tex]
=> [tex]\lambda = 5.723 *10^{-7} \ m[/tex]
This wavelength fall in the range of a yellow light
A mass m = 0.3 kg is released from rest at the origin point 0. The mass falls under the influence of gravity. When the mass reaches point A, it has a velocity of v downward and when the mass reaches point B its velocity is 5v. What is the distance between points A & B divided by the distance between points 0 & A?
Answer:
24
Explanation:
The mass = 3 kg
at point O all the mechanical energy of the system is due to its potential energy PE. The body is at rest.
PE = mgh
but ME = PE + KE = constant (law of energy conservation)
KE is the kinetic energy
since KE is zero at this point, then,
ME = mgh
where m is the mass
g is acceleration due to gravity = 9.81 m/s^2
h is the height = O
ME = 3 x 9.81 x O
ME = 29.43-O
At point A the total ME is due to its PE and its kE
PE at this point = mgh = 3 x 9.81 x A = 29.43-A
KE = [tex]\frac{1}{2}mv^{2}[/tex]
velocity = v at this point, therefore,
KE = [tex]\frac{1}{2}*3*v^{2}[/tex] = [tex]\frac{3}{2} }v^{2}[/tex]
therefore,
ME = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
Equating ME for the points O and A, we have
29.43-O = 29.43-A + [tex]\frac{3}{2} }v^{2}[/tex]
29.43-O - 29.43-A = [tex]\frac{3}{2} }v^{2}[/tex]
(O - A)29.43 = [tex]\frac{3}{2} }v^{2}[/tex]
O - A = 0.051[tex]v^{2}[/tex] this is the distance between point O and A
For point B
PE = 29.43-B
KE = [tex]\frac{25}{2}*3*v^{2}[/tex] = 37.5[tex]v^{2}[/tex] (velocity is equal to [tex]5v[/tex] at this point)
therefore,
ME = 29.43-B + 37.5[tex]v^{2}[/tex]
Equating the ME for points A and B, we have
29.43-A + [tex]\frac{3}{2} }v^{2}[/tex] = 29.43-B + 37.5[tex]v^{2}[/tex]
29.43-A - 29.43-B = 37.5[tex]v^{2}[/tex] - [tex]\frac{3}{2} }v^{2}[/tex]
(A - B)29.43 = 36[tex]v^{2}[/tex]
A - B = 1.22[tex]v^{2}[/tex] this is the distance between points A and B
The distance between points A & B divided by the distance between points 0 & A will be
1.22[tex]v^{2}[/tex]/0.051 = 23.9 ≅ 24
1.
(a)
P
center
Figure 1
A ball is released at point P with a tangential velocity of 5 ms to move in a circular track in a
vertical plane as shown in the Figure 1. Can the ball reach the highest point of the circular track
of radius 1.0 m? Give reasons. (4 marks]
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.
Answer:
137.69°Explanation:
The phase angle of an RLC circuit ϕ is expressed as shoen below;
ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]
Xc is the capacitive reactance = 1/2πfC
Xl is the inductive reactance = 2πfL
R is the resistance = 25.0Ω
Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz
Xl = 2π * 70*0.0940
Xl = 41.32Ω
For the capacitive reactance;
Xc = 1/2π * 70*35.5*10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]
ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]
[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]
Since tan is negative in the 2nd quadrant;
[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]
Hence the phase angle ϕ of the circuit in degrees is 137.69°
The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°
Phase angle:Given that:
capacitance C = 35.5 μF,
Inductance L = 0.0940 H,
The resistance R = 25.0Ω
and frequency f = 70.0Hz
The capacitive reactance is given by:
Xc = 1/2πfC
Xc = 1/2π × 70 × 35.5× 10⁻⁶
Xc = 1/0.0156058
Xc = 64.08Ω
The inductive reactance is given by:
Xl = 2πfL
Xl = 2π × 70 × 0.0940
Xl = 41.32Ω
The phase angle of an RLC circuit ϕ is given by:
[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]
Ф = -42.31°
Since tan is negative in the 2nd quadrant, thus:
ϕ = 180° - 42.31°
ϕ = 137.69°
Learn more about RLC circuit:
https://brainly.com/question/372577?referrer=searchResults
A wooden ice box has a total area of 1.50 m2 amd walls with an average thickness of 2.0 cm. The box contains ice at 0.0 oC. The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the shade of tree at 29 oC. (Assume the thermal conductivity of wood is 0.16 kJ/s m oC
Answer:
m = 9.1 x 10⁶ kg
Explanation:
First, we need to find the rate of heat transfer through the box to the ice. For this purpose, we use Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
Q = Rate Of Heat Transfer = ?
K = Thermal Conductivity = 0.16 KW/m.°C = 160 W/m.°C
A = Area = 1.5 m²
ΔT = Difference in Temperature = 29°C - 0°C = 29°C
L = Thickness of wall = 2 cm = 0.002 m
Therefore,
Q = (160 W/m °C)(1.5 m²)(29°C)/(0.002 m)
Q = 3.48 x 10⁶ W
Now, we find the amount of heat transferred in one day to the ice:
q = Qt
where,
q = amount of heat = ?
t = time = (1 day)(24 h/1 day)(3600 s/1 h) = 86400 s
Therefore,
q = (3.48 x 10⁶ W)(8.64 x 10⁴ s)
q = 3 x 10¹¹ J
Now, for mass of ice melted in a day:
q = m H
m = q/H
where,
m = mass of ice melted in a day = ?
H = latent heat of fusion of ice = 3.3 x 10⁵ J/kg
Therefore,
m = (3 x 10¹¹ J)/(3.3 x 10⁵ J/kg)
m = 9.1 x 10⁶ kg
Please help!
Much appreciated!
Answer:
Rp = 3.04×10² Ω.
Explanation:
From the question given:
1/Rp = 1/4.5×10² Ω + 1/ 9.4×10² Ω
Rp =?
We can obtain the value of Rp as follow:
1/Rp = 1/4.5×10² + 1/ 9.4×10²
Find the least common multiple (lcm) of 4.5×10² and 9.4×10².
The result is 4.5×10² × 9.4×10²
Next, divide the result of the lcm by each denominator and multiply the result obtained with the numerator as shown below:
1/Rp = (9.4×10² + 4.5×10²) /(4.5×10²) (9.4×10²)
1/Rp = 13.9×10²/4.23×10⁵
Cross multiply
Rp × 13.9×10² = 4.23×10⁵
Divide both side by 13.9×10²
Rp = 4.23×10⁵ / 13.9×10²
Rp = 3.04×10² Ω.
A 58 g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g , moves along the x axis at 37 m/s in the positive direction. The second, with mass 22 g , moves along the y axis at 34 m/s in the positive direction. Find the velocity of third piece.
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
With the same block-spring system from above, imagine doubling the displacement of the block to start the motion. By what factor would the following change?
A. Kinetic energy when passing through the equilibrium position.
B. Speed when passing through the equilibrium position.
Answer:
A) K / K₀ = 4 b) v / v₀ = 4
Explanation:
A) For this exercise we can use the conservation of mechanical energy
in the problem it indicates that the displacement was doubled (x = 2xo)
starting point. At the position of maximum displacement
Em₀ = Ke = ½ k (2x₀)²
final point. In the equilibrium position
[tex]Em_{f}[/tex] = K = ½ m v²
Em₀ = Em_{f}
½ k 4 x₀² = K
(½ K x₀²) = K₀
K = 4 K₀
K / K₀ = 4
B) the speed value
½ k 4 x₀² = ½ m v²
v = 4 (k / m) x₀
if we call
v₀ = k / m x₀
v = 4 v₀
v / v₀ = 4
a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?
Answer:
8.78 AmpsExplanation:
Given data:
power rating of the heater P= 1010 W
voltage of the heater V= 115 volts
current taken by the heater I= ?
We can apply the power formula to solve for the current in the heater
i.e P= IV
Making I the current subject of formula we have
I= P/V
Substituting our given data into the expression for I we have
I=1010/115= 8.78 A
Hence the current when the unit/heater is operating is 8.78 AmpAn average sleeping person metabolizes at a rate of about 80 W by digesting food or burning fat. Typically, 20% of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37∘C37 ∘ C, and the skin is typically 7C∘7C ∘ cooler. By how much does the person’s entropy change per second due to this heat transfer?
Answer:
-4.7 x 10^-3 J/K-s
Explanation:
The Power generated by metabolizing food = 80 W
The watt W is equivalent to the Joules per sec J/s
therefor power = 80 J/s
20% of this energy is not used for heating, amount available for heating is
==> H = 80% of 80 = 0.8 x 80 = 64 J/s
The inner body temperature = 37 °C = 273 + 37 = 310 K
The entropy of this inner body ΔS = ΔH/T
ΔS = 64/310 = 0.2065 J/K-s
The skin temperature is cooler than the inner body by 7 °C
Temperature of the skin = 37 - 7 = 30 °C = 273 + 30 = 303 K
The entropy of the skin = ΔS = ΔH/T
ΔS = 64/303 = 0.2112 J/K-s
change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)
==> 0.2065 - 0.2112 = -4.7 x 10^-3 J/K-s
A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is used?
Answer: d₂ = 170 mGya
Explanation:
the relationship between absonbed 'd' and exposure 'E' is given as;
D(Gv) = F . x (AS/xB)
F is a conversion coefficient depending on medium
so we can simply write
d₁/d₂ = x₁/x₂
Given that;
our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?
from the given formula,
d₂ = (x₂d₁ / x₁)
now we substitute
d₂ = (120 × 85) / 60
d₂ = 170 mGya
∴ if 120 mAa is used, the new exposure will be 170 mGya
If two identical wires carrying a certain current in the same direction are placed parallel to each other, they will experience a force of repulsion.
a) true
b) false
Answer:
The answer is B. falseExplanation:
Current in the same direction
When current flow through to parallel conductors of a given length, when the current flows in the same direction
1. A force of attraction between the wires occurs and this tends to draw the wires inward
2. A magnetic field in the same direction is produced.
Current in opposite direction
when the current is in opposite direction
1. Force of repulsion between the two wires occurs, draws the wire outward
2. A magnetic field in opposite direction occurs
If you were to come back to our solar system in 6 billion years, what might you expect to find?
A) a red giant star
B) a rapidly spinning pulsar
C) a white dwarf
D) a black hole
E) Everything will be essentially the same as it is now
Answer:
A)a red giant star
A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero, but the capacitor is charged. If T is the period of the resulting oscillations, the next time, after t = 0 that the energy stored in the magnetic field of the inductor is a maximum is
Answer:
t = T / 2 all energy is stored in the inductor
Explanation:
The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is
w = √1/LC
2π / T =√1 / LC
T = 2π √LC
The energy is constant and for the initial instant it is completely stored in the capacitor
Uc = Q₀² / 2C
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor
U = L I² / 2
in the intermediate instant the energy is stored in the two elements.
Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor
After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,
[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]
here, T is the period of oscillation.
Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,
[tex]U =\dfrac{Q^{2}}{2C}[/tex]
here, C is the capacitance.
In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,
[tex]U'=\dfrac{L I^{2}}{2}[/tex]
here, L is the inductance and I is the current.
Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.
Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation (t = T/2).
Learn more about the capacitance here:
https://brainly.com/question/12644355
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
Answer:
Ok, the question is incomplete buy ill try to answer this in a general way.
Suppose that you have no-polarized light.
When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.
Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°
and:
I(0°) = I0*cos^2(0°) = I0
So the intensity does not change.
Now, knowing the initial intensity, you can find the angle needed to get a given intensity.
For example, if the question was:
"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"
We should solve:
I(θ) = A = I0*cos^2(θ)
(A/i0) = cos^2(θ)
√(A/I0) = cos(θ)
Acos(√(A/I0)) = θ
