Answer:
sinco
Explanation:
A long, current-carrying solenoid with an air core has 1550 turns per meter of length and a radius of 0.0240 m. A coil of 200 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system
Answer:
[tex]M=7.05*10^{-4}[/tex]
Explanation:
From the question we are told that:
Coil one turns N_1=1550 Turns/m
Radius [tex]r=0.0240m[/tex]
Turns 2 [tex]N_2=200N[/tex]
Generally the equation for area is mathematically given by
[tex]A=\pi*r^2[/tex]
[tex]A=\pi*0.024^2[/tex]
[tex]A=\1.81*10^{-3} m^2[/tex]
Therefore
The mutual inductance of this system is
[tex]M=\mu*N_1*N_2*A[/tex]
[tex]M=(4 \pi*10^{-7})*1550*200*1.81*10^{-3}[/tex]
[tex]M=7.05*10^{-4}[/tex]
• Explain how sound travels
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
Sound is a type of energy made by vibrations. These vibrations create sound waves which move through mediums such as air, water and wood. When an object vibrates, it causes movement in the particles of the medium. This movement is called sound waves, and it keeps going until the particles run out of energy.
Suppose a 42-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.125 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.40 T
Answer:
EMF = 73.5 volts
Explanation:
Given that,
The number of a coil, N = 42
The area of the coil, A = 0.125 m²
It is stretched to have no area in 0.100 s
The magnetic field strength is 1.4 T.
We need to find the average induced emf in the coil. We know that,
[tex]\epsilon=N\dfrac{d\phi}{dt}[/tex]
[tex]\epsilon=N\dfrac{BA}{dt}\\\\\epsilon=42\times \dfrac{1.4\times 0.125}{0.1}\\\\=73.5\ V[/tex]
So, the induced emf in the coil is 73.5 volts.
how does the use of standard units of measurement solve problems in measurement regarding validity and reliabiility? explain it
Answer:
Reliability can be estimated by comparing different versions of the same measurement. Validity is harder to assess, but it can be estimated by comparing the results to other relevant data or theory.
An object undergoing simple harmonic motion takes 0.34 s to travel from one point of zero velocity to the next such point. The distance between those points is 27 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answer:
a) T = 0.68 s, b) f = 1.47 Hz, c) λ = 54 m
Explanation:
Simple harmonic motion is described by the expression
x = A cos (wt + Ф)
a) angular velocity and period are related
w = 2π / T
The period is the time it takes for the movement to repeat itself, in this case they indicate the time to reach two consecutive zeros, this corresponds to half a period
T / 2 = 0.34 s
T = 2 0.34
T = 0.68 s
b) period and frequency are inverse
f = 1 / T
f = 1 / 0.68
f = 1.47 Hz
c) the amplitude of the movement is proportional to the energy carried by the wave, which cannot be calculated with the data provided.
With the length data we can find the wavelength
λ / 2 = 27 m
λ = 54 m
What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude
Answer:
The fraction of kinetic energy to the total energy is [tex]\frac{K}{T}=\frac{8}{9}[/tex].
Explanation:
displacement is one third of the amplitude.
Let the amplitude is A.
x= A/3
The kinetic energy of the body executing SHM is
[tex]K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)[/tex]
The total energy is
[tex]T =0.5 mw^2A^2 ..... (2)[/tex]
Divide (1) by (2)
[tex]\frac{K}{T}=\frac{8}{9}[/tex]
A 7.77 kg mass is moving due west at 7.77 m/s. A second mass of 8.88 kg is moving due south at 8.88 m/s. What is the magnitude of their combined velocity, in m/s, after they collide and stick together?
a) 1.11 m/s
b) 3.82 m/s
c) 5.96 m/s
d) 11.8 m/s
Answer:
8.362m/s
Explanation:
Given data
Mass m1= 7.77kg
Velocity v1= 7.77m/s
Mass m2= 8.88kg
Velocity v2= 8.88m/s
Apply the law of conservation of momentum for inelastic collision we have
m1v1+m2v2= (m+m2)V
7.77*7.77+ 8.88*8.88= (7.77+8.88)V
60.3729+78.8544= 16.65V
139.2273= 16.65V
Divide both sides by 16.65
V= 139.2273/16.65
V= 8.362m/s
Hence the final velocity is 8.362m/s
Nhiệt dung riêng của một chất là ?
Answer:
enchantment table language
Explanation:
A bus is moving at a speed of 20 m/s, when it begins to slow at a constant rate of 5 m/s2 in order to stop at a bus stop. If it comes to rest at the bus stop, how far away was the bus from the stop?
Answer:
40 m
Explanation:
Applying,
v² = u²+2as............... Equation 1
Where v = final velocity, u = initial velocity, a = deceleration, s = distance
make s the subject of the equation
s = (v²-u²)/2a............ Equation 2
From the question,
Given: v = 0 m/s (comes to rest), u = 20 m/s, a = -5 m/s² (begins to slow)
Substitute these values into equation 2
s = (0²-20²)/[2(-5)]
s = -400/-10
s = 40 m
Hence the bus was 40 m from the stop
Two distinct systems have the same amount of stored internal energy. 500 J are added by heat to the first system and 300 J are added by heat to the second system. What will be the change in internal energy of the first system if it does 200 J of work? How much work will the second system have to do in order to have the same internal energy?
Answer:
The change in the internal energy of the first system is 300 J
The second system will do zero work in order to have the same internal energy.
Explanation:
Given;
heat added to the first system, Q₁ = 500 J
heat added to the second system, Q₂ = 300 J
work done by the first system, W₁ = 200 J
The change in the internal energy of the system is given by the first law of thermodynamics;
ΔU = Q - W
where;
ΔU is the change in internal energy of the system
The change in the internal energy of the first system is calculated as;
ΔU₁ = Q₁ - W₁
ΔU₁ = 500 J - 200 J
ΔU₁ = = 300 J
The work done by the second system to have the same internal energy with the first.
ΔU₁ = Q₂ - W₂
W₂ = Q₂ - ΔU₁
W₂ = 300 J - 300 J
W₂ = 0
The second system will do zero work in order to have the same internal energy.
Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given point in the field is 80 V, what is the voltage at a point 1.0 m directly east of the point
Answer:
30 V
Explanation:
Given that:
The uniform electric field = 50 N/C
Voltage = 80 V
distance = 1.0 m
The potential difference of the electric field = Δ V
E_d = V₁ - V₂
50 × 1 = 80V - V₂
50 - 80 V = - V₂
-30 V = - V₂
V₂ = 30 V
The distance between the two object is fixed at 5.0 m. The uncertainty distance measurement is? The percentage error in the distance is?
Physicists and engineers from around the world have come together to build the largest accelerator in the world, the Large Hadron Collider (LHC) at the CERN Laboratory in Geneva, Switzerland. The machine accelerates protons to high kinetic energies in an underground ring 27 km in circumference.
a. What speed v of proton in the LHC if the proton's kinetic energy is 6.7 TeV? (Because v is very close to c, write v=(1−Δ)c and give your answer in terms of Δ).
b. Find the relativistic mass, mrel, of the accelerated protons in terms of their rest mass.
Solution :
Energy of photon, E = 6.7 eV
E = [tex]$6.7 \times 1.602 \times 10^{-7}$[/tex] joule
Kinetic energy, [tex]$K.E. =\frac{1}{2} mv^2 = 1.602 \times 6.7 \times 10^{-7}$[/tex]
[tex]$v^2=\frac{2 \times 1.602 \times 6.7 \times 10^{-7}}{1.6726 \times 10^{-27}}$[/tex]
[tex]$=12.834 \times 10^{-20}$[/tex]
Kinetic energy at high speeds
[tex]$(r-1)\times mc^2 = 6.7 \ eV$[/tex]
[tex]$(r-1)=\frac{6.7 \times 1.602 \times 10^{-7}}{1.6726 \times 10^{-27} \times 9 \times 10^{16}}$[/tex]
r - 1 = 7130
r = 7130 + 1
r = 7131
[tex]$\frac{1}{\sqrt{1-\frac{v^2}{C^2}}}=7131$[/tex]
[tex]$1-\frac{v^2}{C^2} = \left(\frac{1}{7131}\right)^2$[/tex]
[tex]$v^2=C^2\left[1-\left(\frac{1}{7131}\right)^2\right]$[/tex]
[tex]$v=0.99999999017C$[/tex]
Δ = 1 - 0.99999999017
= 0.00000000933
Relative mass, [tex]$m_{rel}=r.m$[/tex]
[tex]$=7131 \times 1.6728 \times 10^{-27}$[/tex]
[tex]$=1.1927 \times 10^{-23}$[/tex] kg
A kugel fountain, also known as a floating sphere fountain, is a sculpture that suspends a large, typically stone, sphere on a thin plane of water. One of three in Columbus, is on campus in the Chadwick Arboretum on the corner of Lane Ave. and Olentangy. Kugel fountains have very little friction between the stone sphere and the hollow in which they sit, so it takes very little torque to start one moving from rest.
Required:
a. The kugel fountain in the Richard and Annette Bloch Cancer Survivor's Plaza in the Lane Ave Gardens has a granite sphere with a mass of approximately 1,800 kg, and a radius of about half a meter. You use your hand to apply a tangential force on the sphere to start it rotating from rest, and it takes 12 seconds to complete a quarter revolution. What is the average torque you applied to the sphere?
b. What is the average magnitude of the tangential force you applied?
c. How does this compare to the weight of a cellphone?
Answer:
Explanation:
Moment of inertia of sphere I = 2 /5 x m R²
m is mass and R is radius of the sphere .
I = 2/5 x 1800 x 0.5²
= 180 kg m².
For rotational motion starting from rest
θ = 1/2 α t²
Here θ = π /2 = 1.57 radians
t = 12 s
1.57 radians = 1/2 x α x 12²
Angular acceleration α = .022 rad /s²
Torque applied = I α
= 180 kg m² x .022 rad /s²
= 3.96 N.m
b ) Torque = F x r
F = Torque / r
= 3.96 / .5 = 7.92 N .
c )
Average weight of a cellophone is 1.3 N .
So the force applied is a little more than weight of a cellophone.
A proposed space station includes living quarters in a circular ring 51.0 m in diameter. At what angular speed should the ring rotate so the occupants feel that they have the same weight as they do on Earth?
Answer:
0.620 rad/s
Explanation:
Given that:
diameter of the ring = 51.0 m
radius = diameter/2 = 51.0 m/2
= 25.5 m
In the system, the centripetal force is equal to the force as a result of the weight;
∴
mω²r = mg
ω²r = g
where:
ω = angular speed
[tex]\omega = \sqrt{\dfrac{g}{r}} \\ \\ \omega = \sqrt{\dfrac{9.8}{25.5}}[/tex]
ω = 0.620 rad/s
A wheel rotates at an angular velocity of 30rad/s. If an acceleration of 26rad/s2 is applied to it, what will its angular velocity be after 5.0s
g A CD is spinning on a CD player. You open the CD player to change out the disk and notice that the CD comes to rest after 15 revolutions with a constant deceleration of 120 r a d s 2 . What was the initial angular speed of the CD
Answer:
[tex]\omega_1=150rads/sec[/tex]
Explanation:
From the question we are told that:
Number of Revolution [tex]N=15=30\pi[/tex]
Deceleration [tex]d= -120 rads/2[/tex]
Generally the equation for initial angular speed [tex]\omega_1[/tex] is mathematically given by
[tex]\omega_2^2=\omega_1^2 +2(d)(N)[/tex]
[tex]0=\omega_1^2 +2(-120)(20 \pi)[/tex]
[tex]\omega_1^2=7200 \pi[/tex]
[tex]\omega_1=150rads/sec[/tex]
Welcome to this IE. You may navigate to any page you've seen already using the IE Outline tab on the right. A particle beam is made up of many protons each with a kinetic energy of 3.25 x 10-15J. A proton has a mass of 1.673x10-27kg and a charge of 1.602x10-19C. What is the magnitude of a uniform electric field that will stop these protons in a distance of 2 m
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, what correctly describes the path difference of the two beams
Answer:
Two beams (that we could think as sinusoidal waves) face destructive interference at a point P, if at that point one of the waves is in a peak, and the other wave is in a through, so when we add them, the waves will "cancel" each other (thus, we have destructive interference)
The case of constructive interference happens when at point P we have two peaks or two throughs, so the waves add up.
If both waves started at the same point and with the same phase and both traveled the same distance, they will always have constructive interference. But if the waves traveled different distances, the constructive interference will happen at points where the path difference of the waves is an integer multiple of the wavelength (remember that the distance between a peak and the next one is the wavelength).
If two beams of coherent light start at the same point and are in phase, then we will have the maximum of destructive interference at a point P if the path difference is exactly half of the wavelength (or (n/2) times the wavelength, with n an odd integer), this happens because the distance between a peak and the next thtough is exactly half of the wavelength.
Then we can conclude that the path difference between the two beams is of the form:
(n/2)*λ
where:
n = odd number = (2*k + 1) with k an integer.
λ = wavelength of the beams.
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Frisbee. Ignoring air resistance, the torque exerted about its center of mass by gravity is: __________
a. 0.
b. mgr
c. 2mgr
d. a function of the angular velocity.
e. small at first, then increasing as the Frisbee loses the torque given it by your hand.
Answer:
the correct answer is a
Explanation:
The torque is
τ = F x r
where the bold letters indicate vectors, in this case the vector of the center of mass is perpendicular to the weight of the body
τ = mg r
in body weight it is applied at the point of the center of mass, therefore as the distance of the force from the axis of rotation (center of amas) is zero, the die is zero
the correct answer is a
1. A positive electric charge in a region of changing electric potential will: A. move in the direction of decreasing potential B. move in the direction of increasing potential C. move in an undetermined way; we need more information D. remain at rest
Answer:
The correct option is (B).
Explanation:
The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field. The electric potential due to a point charge is given by :
[tex]V=\dfrac{kQ}{r}[/tex]
Where
k is the electrostatic constant
Q is the electric charge
r is the distance from the charge
So, a positive electric charge in a region of changing electric potential will move in the direction of increasing potential.
A crate with a mass of 161.5 kg is suspended from the end of a uniform boom with a mass of 72.3 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall. Calculate the tension in the cable.
Answer:
The correct answer is "2205.72 N".
Explanation:
Given:
m₁ = 161.5 kg
m₂ = 72.3 kg
By taking moment about point A, we get
⇒ [tex]m_1 g Cos \theta.\frac{l}{2 Cos \theta} + m g Cos \theta.\frac{l}{ Cos \theta} -T Cos(90-2 \theta).\frac{l}{2 Cos \theta} = 0[/tex]
By substituting the values, we get
⇒ [tex]\frac{161.5\times 9.8\times 13}{2}+72.3\times 9.8\times 13-T.2 Sin \theta.l =0[/tex]
⇒ [tex]10287.55+9211.02-T 2\times 0.34\times 13=0[/tex]
⇒ [tex]T\times 8.84=19498.57[/tex]
⇒ [tex]T= 2205.72 N[/tex]
An artificial satellite circling the Earth completes each orbit in 125 minutes. (a) Find the altitude of the satellite.(b) What is the value of g at the location of this satellite?
Answer:
(a) Altitude = 1.95 x 10⁶ m = 1950 km
(b) g = 5.9 m/s²
Explanation:
(a)
The time period of the satellite is given by the following formula:
[tex]T^2 = \frac{4\pi^2r^3}{GM_E}[/tex]
where,
T = Time period = (125 min)([tex]\frac{60\ s}{1\ min}[/tex]) = 7500 s
r = distance of satellite from the center of earth = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
[tex]M_E[/tex] = Mass of Earth = 6 x 10²⁴ kg
Therefore,
[tex](7500\ s)^2 = \frac{4\pi^2r^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}\\\\r^3 = \frac{(7500\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{5.7\ x\ 10^{20}\ m^3} \\[/tex]
r = 8.29 x 10⁶ m
Hence, the altitude of the satellite will be:
[tex]Altitude = r - radius\ of\ Earth\\Altitude = 8.29\ x\ 10^6\ m - 6.34\ x\ 10^6\ m[/tex]
Altitude = 1.95 x 10⁶ m = 1950 km
(b)
The weight of the satellite will be equal to the gravitational force between satellite and Earth:
[tex]Weight = Gravitational\ Force\\\\M_sg = \frac{GM_EM_s}{r^2}\\\\g = \frac{GM_E}{r^2}\\\\g = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{(8.23\ x\ 10^6\ m)^2}[/tex]
g = 5.9 m/s²
Physics question plz help ASAP
Hello everyone.Roads are made winding in hilly regions why?Explain
Answer:
Because winding roads have a gentle slope on hills, so it's easy to climb it than a steepy.
In a calorimetry experiment, three samples A, B, and C with TA> TB> Tc are placed in thermal contact. When the samples have reached thermal equilibrium at a common temperature T, which one of the following must be true?
a. QA > QB >QC
b. QA< 0, QB <0, and Qc > 0
c. T> TB
d. T
e. TA > T> Tc
Answer:
e. TA>T>Tc
Explanation:
a) In this case, we cannot say for sure QA>QB>QC. This is because the magnitude of the heat flow will depend on the specific heat and the mass of each sample. Due to the equation:
[tex]Q=mC_{p}(T_{f}-T_{0})[/tex]
if we did an energy balance of the system, we would get that>
QA+QB+QC=0
For this equation to be true, at least one of the heats must be negative. And one of the heats must be positive.
We don't know either of them, so we cannot determine if this statement is true.
b) We can say for sure that QA<0, because when the two samples get to equilibrum, the temperatrue of A must be smaller than its original temperature. Therefore, it must have lost heat. But we cannot say for sure if QB<0 because sample B could have gained or lost heat during the process, this will depend on the equilibrium temperature, which we don't know. So we cannot say for sure this option is correct.
c) In this case we don't know for sure if the equilibrium temperature will be greater or smaller than TB. This will depend on the mass and specific heat of the samples, just line in part a.
d) is not complete
e) We know for sure that A must have lost heat, so its equilibrium temperature must be smaller than it's original temperature. We know that C must have gained heat, therefore it's equilibrium temperature must be greater than it's original temperature, so TA>T>Tc must be true.
Can an electron be diffracted? Can it exhibit interference?
Answer:
Yeah, it can be diffracted. Though it depends on a diffracting medium.
It must have some magnetic fields .
Forexample; X-ray diffraction where electrons are diffracted to the target filament.
A building is being knocked down with a wrecking ball, which is a big metal sphere that swings on a 15-m-long cable. You are (unwisely!) standing directly beneath the point from which the wrecking ball is hung when you notice that the ball has just been released and is swinging directly toward you. How much time do you have to move out of the way? answer in seconds.
Answer:
Time to move out of the way = 1.74 s
Explanation:
Time to move out of the way is one fourth of period = 6.95/4 = 1.74 seconds.
Time to move out of the way = 1.74 s
A monochromatic light falls on two narrow slits that are 4.50 um apart. The third destructive fringes which are 35° apart were formed at a screen 2m from the light source. The light source is 0.30 m from the slits. () Calculate Ym. (4 marks) Compute the wavelength of the light. (4 marks)
Answer:
y = 1.19 m and λ = 8.6036 10⁻⁷ m
Explanation:
This is a slit interference problem, the expression for destructive interference is
d sin θ = m λ
indicate that for the angle of θ = 35º it is in the third order m = 3 and the separation of the slits is d = 4.50 10⁻⁶ m
λ = d sin θ / m
let's calculate
λ = 4.50 10⁻⁶ sin 35 /3
λ = 8.6036 10⁻⁷ m
for the separation distance from the central stripe, we use trigonometry
tan θ= y / L
y = L tan θ
the distance L is measured from the slits, it indicates that the light source is at x = 0.30 m from the slits
L = 2 -0.30
L = 1.70 m
let's calculate
y = 1.70 tan 35
y = 1.19 m
The USDA recommends that women consume about 2000 Calories per day and men consume about 2500 Calories per day. How much average power use (in Watts) does this imply for a human body
Answer:
1500 W to 2200 W
Explanation:
Every person does work in his daily day to day life. A person needs energy in order to perform work. And the energy consumed by an individual while performing a daily work is directly responsible to the mount of oxygen consumed by the person.
The USDA is the federal agency which looks after the food and agriculture matters of the US government. It deals with and formulates different policies and laws for the country and it s people. It recommends about 2000 calories per day for women and for men it recommends about 2500 calorie per days of food intake.
Accordingly, the average power required by a human body for doing regular work is in the range of 1500 W to 2200 W.
