Explanation:
According to Thermal Expansion of solids:
[tex]dl = \alpha \times l \times dt[/tex]
[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]
[tex]dl = 7.2 \times {10}^{ - 3} [/tex]
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?
Answer:
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Explanation:
When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.
In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:
[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]
Where:
[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.
If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:
[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]
[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]
And change in heat energy ([tex]Q[/tex]), measured by joules, by:
[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]
Where:
[tex]n[/tex] - Molar quantity, measured in moles.
The final temperature of the monoatomic ideal gas is now cleared:
[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]
Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:
[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]
[tex]T_{2} = 420.279\,K[/tex]
The final pressure of the system is calculated by the following relationship:
[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]
If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:
[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]
[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.165×105 km and the planet has a radius of 4175 km and a mass of 6.70×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2 .
Answer:
364days
Explanation:
Pls see attached file
Explanation:
The moon will take 112.7 days to make one revolution around the planet.
What is Kepler's third law?The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.
Given, the distance from the moon's center to the planet's surface,
h = 2.165 × 10⁵ km,
The radius of the planet, r = 4175 km
The mass of the planet = 6.70 × 10²² kg
The total distance between the moon's center to the planet's center:
a = r +h = 2.165 × 10⁵ + 4175
a = 216500 + 4175
a = 220675
a = 2.26750 × 10⁸ m
The period of the planet can be calculated as:
[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]
[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]
T = 9738253.26 s
T = 112.7 days
Learn more about Kepler's law, here:
https://brainly.com/question/1608361
#SPJ5
The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be
Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]
or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
The elastic limit of an alloy is 5.0×108 N/m2. What is the minimum radius rmin of a 4.0 m long wire made from the alloy if a single strand is designed to support a commercial sign that has a weight of 8000 N and hangs from a fixed point? To stay within safety codes, the wire cannot stretch more than 5.0 cm.
Answer:
4.5x 10^ -9m
Explanation:
See attached file
Answer:
The radius is [tex]r_{min} = 0.00226 \ m[/tex]
Explanation:
From the question we are told that
The elastic limit(stress) is [tex]\sigma = 5.0*10^{8} \ N /m^2[/tex]
The length is [tex]L = 4.0 \ m[/tex]
The weight of the commercial sign is [tex]F_s = 8000 \ N[/tex]
The maximum extension of the wire is [tex]\Delta L = 5.0 \ cm = 0.05 \ m[/tex]
Generally the elastic limit of an alloy (stress) is is mathematically represented as
[tex]\sigma = \frac{ F_s }{ A }[/tex]
Where A is the cross-sectional area of the wire which is mathematically represented as
[tex]A = \pi r^2[/tex]
here [tex]r = r_{min}[/tex] which is the minimum radius of the wire that support the commercial sign
So
[tex]\sigma = \frac{ F_s }{ \pi r_{min}^2 }[/tex]
=> [tex]r_{min} = \sqrt{\frac{F_s}{\sigma * \pi} }[/tex]
substituting values
[tex]r_{min} = \sqrt{\frac{8000}{ 5.0* 10^8 * 3.142} }[/tex]
[tex]r_{min} = 0.00226 \ m[/tex]
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleration does it experience
Answer:
The angular acceleration is [tex]\alpha = 3.235 \ rad/s ^2[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.034\ kg \cdot m^2[/tex]
The net torque is [tex]\tau = 0.11\ N \cdot m[/tex]
Generally the net torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where [tex]\alpha[/tex] is the angular acceleration so
[tex]\alpha = \frac{\tau }{I}[/tex]
substituting values
[tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]
[tex]\alpha = 3.235 \ rad/s ^2[/tex]
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
A city of punjab has 15 percemt chance of wet weather on any given day. What is probability that it will take a week for it three wet weather on 3 sepaprate days? Also find it standard deviation
Answer:
The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
Explanation:
We are given that A city of Punjab has 15 percent chance of wet weather on any given day.
So, Probability of wet weather = 0.15
Probability of not being a wet weather = 1-0.15 =0.85
We are supposed to find probability that it will take a week for it three wet weather on 3 separate days
Total number of days in a week = 7
We will use binomial over here
n = 7
p =probability of failure = 0.15
q = probability of success=0.85
r=3
Formula :[tex]P(r=3)=^nC_r p^r q ^{n-r}[/tex]
[tex]P(r=3)=^{7}C_{3} (0.15)^3 (0.85)^{7-3}\\P(r=3)=\frac{7!}{3!(7-3)!} (0.15)^3 (0.85)^{7-3}\\P(r=3)=0.06166[/tex]
Standard deviation =[tex]\sqrt{n \times p \times q}[/tex]
Standard deviation =[tex]\sqrt{7 \times 0.15 \times 0.85}[/tex]
Standard deviation =0.9447
Hence The probability that it will take a week for it three wet weather on 3 separate days is 0.06166 and its standard deviation is 0.9447
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm
Answer:
The maximum force that can be lifted by this system is 51,478.4 kg
Explanation:
Given;
maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²
diameter of the piston, d = 65 cm = 0.65 m
The maximum gauge pressure of the piston is given as;
[tex]Hp = \frac{F}{A}[/tex]
Where;
F is the maximum force of the piston
A is the area of the piston
[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]
F = Hp x A
F = 1.52 x 10⁶N/m² x 0.3319m²
F = 504488 N
Force is given as;
F = mg
m = F/g
m = 504488/9.8
m = 51,478.4 kg
Therefore, the maximum force that can be lifted by this system is 51,478.4 kg
An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum
Explanation:
The De-Broglie wavelength is given by :
[tex]\lambda=\dfrac{h}{p}[/tex]
h is Planck's constant
p is momentum
In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.
Only momentum is the factor that remains same for both particles i.e. momentum.
An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for the blood in an artery during heart surgery). Such a device is illustrated. The conducting fluid moves with velocity v in a tube of diameter d perpendicular to which is a magnetic field B. A voltage V is induced between opposite sides of the tube. Given B = 0.120 T, d = 1.2 cm., and a measured voltage of 2.88 mV, determine the speed of the blood.
Answer:
2 m/s
Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as
[tex]E = Blv[/tex]
where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
[tex]v[/tex] is the velocity of the fluid through the field = ?
[tex]B[/tex] is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]
2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]
[tex]v[/tex] = 2.88/1.44 = 2 m/s
A ball is thrown from the ground so that it’s initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Find the ball’s total time of flight and distance it traverses before hitting the ground.
Answer:
8 seconds
160 meters
Explanation:
Given in the y direction:
Δy = 0 m
v₀ = 40 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (40 m/s) t + ½ (-10 m/s²) t²
0 = 40t − 5t²
0 = 5t (8 − t)
t = 0 or 8
Given in the x direction:
v₀ = 20 m/s
a = 0 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²
Δx = 160 m
Your favorite radio station broadcasts at a frequency of 91.5 MHz with a power of 11.5 kW. How many photons does the antenna of the station emit in each second?
Answer:
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
Explanation:
Given:
Frequency = 91.5 MHz
Power = 11.5 Kw = 11,500 J/s
Find:
Number of photons emit per second
Computation:
Total energy with frequency (E) = hf
Total energy with frequency (E) = 6.626×10⁻³⁴ × 91.5×10⁶
Total energy with frequency (E) = 6.06×10⁻²⁶ J
Number of photons emit per second = 11,500 / 6.06×10⁻²⁶
Number of photons emit per second = 1897.689 × 10²⁶
Number of photons emit per second = 1.9 × 10²⁹ (Approx)
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?
Answer: 0.0617
Explanation:
Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15
Let X be a binomial variable that represents the number of days having wet weather.
Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.
Here, n= 7 ( 1 week = 7 days)
The probability that it will take a week for it three wet weather on 3 separate days:
[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]
Hence, the required probability =0.0617
The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
Near Point(cm) Far Point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]
Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.
1. Berenice
2. Avishka
3. Francesca
4. Edouard
Answer:
1. Berenice = 0.67 D
2. Avishka = 1.50 D
3. Francesca = 2.00 D
4. Edouard = 2.75 D
Explanation:
The farsighted people are those with near point greater than 25 cm.
They include Avishka, Berenice, Edouard and Francisca.
A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative
From the lens formula, 1/f = 1/v 1/u; but v is negative
Therefore, 1/f = 1/u - 1/v
But, power of a lens = 1/f in meters.
Therefore, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
P = 1/ 0.25 - 1/0.4 = 1.5 D
For Berenice, u = 0.25 m, v = 0.3 m
P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v = 0.80 m
P = 1/0.25 - 1/0.80 =2.75 D
For Francesca, u = 0.25 m, v = 0.50 m
P = 1/0.25 - 1/0.5 = 2.0 D
When The farsighted people are those with a near point greater than 25cm.
Berenice is = 0.67 D
Avishka is = 1.50 D
Francesca is = 2.00 D
Edouard is = 2.75 D
What is Hyperopic Vision?
When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.
Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.
Also, From the lens formula, 1/f = 1/v 1/u; but v is negative
Thus, 1/f = 1/u - 1/v
But, when the power of a lens = 1/f in meters.
Thus, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
After that, P = 1/ 0.25 - 1/0.4 = 1.5 D
Then, For Berenice, u = 0.25 m, v = 0.3 m
Now, P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v is = 0.80 m
Then, P = 1/0.25 - 1/0.80 is =2.75 D
Now, For Francesca, u = 0.25 m, v is = 0.50 m
Therefore, P = 1/0.25 - 1/0.5 = 2.0 D
Find more information about Hyperopic Vision here:
https://brainly.com/question/25676535
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth
Answer:
g' = g/9 = 1.09 m/s²
Explanation:
The magnitude of free fall acceleration at the surface of earth is given by the following formula:
g = GM/R² ----- equation 1
where,
g = free fall acceleration
G = Universal Gravitational Constant
M = Mass of Earth
R = Distance between the center of earth and the object
So, in our case,
R = R + 2 R = 3 R
Therefore,
g' = GM/(3R)²
g' = (1/9) GM/R²
using equation 1:
g' = g/9
g' = (9.8 m/s)/9
g' = 1.09 m/s²
Answer:
The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]Explanation:
Surface of earth,
[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]
free fall acceleration at height h,
[tex]g_h = \frac{GM}{(R+h)^2}[/tex]
where
G = gravitational constant
R = Radius of earth
M = mass of earth
therefore,
[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]
Where height h = 2R
[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]
For more information, visit
https://brainly.com/question/17162343
Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided
Answer:
R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor
Explanation:
For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)
Therefore we build a circuit with 4 resistors in series, in such a way that
V = i R
let the voltage
1st resistance
V = i R
R₁ = V / i
R₁ = 14.7 / 1 10⁻³
R₁ = 14.7 10³ Ω
power is
P = V i
P = 14.7 1 10⁻³
P = 14.7 10⁻³ W = 0.0147 W
a resistance of ⅛ W is indicated
2nd resistance
R₂ = 8.18 / 1 10⁻³
R₂ = 8.18 10³ Ω
Power
P = 8.18 1 10⁻³
P = 0.00818W
a 1/8 W resistor
3rd resistance
this resistance is calculated in such a way that
V₁ + V₂ + V₃ = 24.6
V₃ = 24.6 - V₁ -V₂
V₃ = 24.6 - 14.7 - 8.18
V₃ = 1.72 V
R₃ = 1.72 / 1 10⁻³
R₃ = 1.72 10³ Ω
power
P = Vi
P = 1.72 10⁻³
P = 0.00172 W
a resistance of ⅛ W
To obtain the voltage of 24.6 we use this three resistors together
4th resistance
The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage
30 = V₁ + V₂ + V₃ + V₄
V₄ = 30 - V₁ -V₂ -V₃
V₄ = 30 -14.7 - 8.18 - 1.72
V₄ = 5.4 V
R₄ = 5.4 / 1 10⁻³
R₄ = 5.4 10³ Ω
Power
P = V i
P = 5.4 10⁻³
P = 0.0054 W
⅛ W resistance
The values of these resistance are commercially
Let's check the consumption of the circuit
R_total = R₁ + R₂ + R₃ + R₄
R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³
R_total = 30 10³
the current circulating in the circuit is
i = V / R_total
i = 30/30 10³
i = 1 10⁻³ A
therefore it is within the order requirement.
for connections see attached diagram
At the pizza party you and two friends decide to go to Mexico City from El Paso, TX where y'all live. You volunteer your car if everyone chips in for gas. Someone asks how much the gas will cost per person on a round trip. Your first step is to call your smarter brother to see if he'll figure it out for you. Naturally he's too busy to bother, but he does tell you that it is 2015 km to Mexico City, there's 11 cents to the peso, and gas costs 5.8 pesos per liter in Mexico. You know your car gets 21 miles to the gallon, but we still don't have a clue as to how much the trip is going to cost (in dollars) each person in gas ($/person).
Answer:
cost_cost = $ 96
Explanation:
In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system
performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)
perfomance= 8,927 km / l
now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km
#_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters
Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos
cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638
cost_gasoline = #_gasoline cost_litro
cost_gasoline = 225.72 0.638
cost_gasoline = $ 144
This cost is for the one way trip, the total round trip cost is
cost_total = 2 cost_gasoline
cost_total = $ 288
Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is
cost_person = total_cost / #_people
cost_person = 288/3
cost_cost = $ 96
A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.
Answer:
The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Explanation:
Given that,
A beam of light from a laser illuminates a glass.
Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]
Index of refraction is 2.83.
We need to calculate the speed of light pulse in glass
Using formula of speed
[tex]v=\dfrac{c}{\mu}[/tex]
Put the value into the formula
[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]
[tex]v=1.06\times10^{8}\ m/s[/tex]
We need to calculate the time of short pulse of light beam
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]
[tex]t=2.37\times10^{-9}\ sec[/tex]
Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly
Answer:
λ = 667.85 nm
Explanation:
Let f be the frequency detected by the observer
Let v be the speed at which the observer is moving.
Now, when the direction at which the observer is moving is away from the source, we have the frequency as;
f = f_o√((1 - β)/(1 + β))
From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;
λ = λ_o√((1 + β)/(1 - β))
Where, β = Hr/c
H is hubbles constant which has a value of 0.0218 m/s • ly
c is speed of light = 3 × 10^(8) m/s
r is given as 2.40 x 10^(8) ly
Thus,
β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))
β = 0.01744
Since we are given λ_o = 656.3 nm
Then;
λ = 656.3√((1 + 0.01744)/(1 - 0.01744))
λ = 667.85 nm
When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?
Answer:
The net force is directed downwards.
Explanation:
Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.
