Answer:[tex]3800\ W[/tex]
Explanation:
Given
Lawn mover running for [tex]t=20\ min [/tex]
and does [tex]W=4560\times 10^3\ J[/tex]
We know Power is rate of work i.e.
[tex]P=\frac{\text{Work}}{\text{time}}[/tex]
[tex]P=\frac{4560\times 10^3}{20\times 60}[/tex]
[tex]P=3800\ W[/tex]
Thus Power output is [tex]3800\ W[/tex]
How does a thermos keep hot drinks hot?
A: By absorbing radiant energy that might enter the thermos
B: By limiting the amount of thermal energy that flows out of the drink
C: By limiting the amount of cold that flows into the drink
D: By transmitting radiant energy into the thermos
Answer:
the answer is B
Explanation:
hope this helps
Answer:
b
Explanation:
can someone help asap with this
Answer:
B
Explanation:
i had a test on this and got it correct
4. An electric toaster is rated at 500 watts. Determine the amount of electrical
energy it converts to heat energy in one minute.
80.3 J
B)
500 J
C)
3,000 J
D)
30,000 J
Answer:60000 joules
Explanation:
power=500watts
Time=2minutes=120seconds
Heat energy=power x time
Heat energy=500 x 120
Heat energy=60000J
An electric toaster rated at 500 watts converts electrical energy into heat energy at a rate of 500 joules per second. To find the energy used in one minute, we simply multiply the wattage by the seconds in a minute, resulting in 30,000 joules.
Explanation:An electric toaster rated at 500 watts converts electrical energy into heat energy at a rate of 500 joules per second. Since watts are joules per second, to find out how much energy you would use in one minute, you need to multiply the number of watts by the number of seconds in one minute. There are 60 seconds in a minute so the computation is 500 joules/second × 60 seconds = 30000 joules. Therefore, in one minute, a 500-watt toaster will convert 30,000 joules of electrical energy into heat energy.
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As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a channel in a microprocessor. If we know that an electron is somewhere along the 50 nm length of the channel, what is ∆vx? If we treat the elec- tron as a classical particle moving at a speed at the outer edge of the uncertainty range, how long would it take to traverse the channel?
Answer:
a) ∆x∆v = 5.78*10^-5
∆v = 1157.08 m/s
b) 4.32*10^{-11}
Explanation:
To solve this problem you use the Heisenberg's uncertainty principle, that is given by:
[tex]\Delta x\Delta p \geq \frac{\hbar}{2}[/tex]
where h is the Planck's constant (6.62*10^-34 J s).
If you assume that the mass of the electron is constant you have:
[tex]\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}[/tex]
you use the value of the mass of an electron (9.61*10^-31 kg), and the uncertainty in the position of the electron (50nm), in order to calculate ∆x∆v and ∆v:
[tex]\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s[/tex]
[tex]\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}[/tex]
If the electron is a classical particle, the time it takes to traverse the channel is (by using the edge of the uncertainty in the velocity):
[tex]t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s[/tex]
A conducting wire formed in the shape of a right triangle with base b = 34 cm and height h = 77 cm and having resistance R = 2.2 Ω, rotates uniformly around the y-axis in the direction indicated by the arrow (clockwise as viewed from above (loooking down in the negative y-direction)). The triangle makes one complete rotaion in time t = T = 1.6 seconds. A constant magnetic field B = 1.6 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating.1)What is ?, the angular frequency of rotation?2)What is Imax, the magnitude of the maximum induced current in the loop?3)At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux ?1 at time t = t1 = 0.45 s?4)What is I1, the induced current in the loop at time t = 0.45 s? I1 is defined to be positive if it flows in the negative y-direction in the segment of length h.
Answer:
a) 3.92 rad/s
b) 0.373 A
d) 0.018 A
Explanation:
a) The angular frequency of rotation is given by:
[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{(1.6s)}=3.92\frac{rad}{s}[/tex]
b) The maximum induced current in the loop is given by:
[tex]I_{max}=\frac{emf_{max}}{R}=\frac{AB\omega}{R}[/tex]
R: resistance
A: area of the triangle loop = bh/2 = (0.34m)(0.77m)/(2) = 0.1309m^2
B: magnitude of the magnetic field
[tex]I_{max}=\frac{(0.13m^2)(1.6T)(3.92rad/s)}{2.2\Omega}=0.373A[/tex]
d) For t = 0.45s you have:
[tex]I(t)=\frac{ABcos(\omega t)}{R}\\\\I(0.45)=\frac{(0.13m^2)(1.6T)cos(3.92rad/s \ (0.45s))}{2.2\Omega}=-0.018A[/tex]
But I1 is defined to be positive if it flows in the negative y-direction.
hence, I for t=0.45 s is 0.018A
(1) The angular frequency is 3.92 rad/s
(2) the value of I(max) = 0.373A
(3) Image is required
(4) The current at 0.45s is -0.018A
Induced current:Given a conducting wire formed in the shape of a right triangle with:
Base b = 34 cm
Height h = 77 cm
Resistance R = 2.2 Ω
(1) Time period of rotation is T = 1.6s
The angular frequency is given by:
ω = 2π/T
ω = 2π/1.6
ω = 3.92 rad/s
(2) The magnetic field applied is B = 1.6T, perpendicular to the plane of the triangle.
the induced current is given by:
[tex]I_{ind}=\frac{EMF_{ind}}{R}[/tex]
where R is the resistance
[tex]I_{max}=\frac{EMF_{max}}{R}\\\\I_{max}=\frac{BA\omega}{R}\\\\I_{max}=\frac{1.6\times(1/2)\times34\times77\times3.92}{2.2}\\\\[/tex]
I(max) = 0.373A
(3) The image for the question is not attached.
(4) The instantaneous value of induced current at time t = 0.45s is given by:
[tex]I_{ind}=\frac{EMF_{ind}}{R}\\\\I(t)=\frac{BA(cos\omega t)}{R}\\\\I(t)=\frac{(1/2)\times34\times77\times1.6\timescos(3.92\imes0.45)}{2.2}[/tex]
I(t) = -0.018 A
the negative sign indicated that the current flows in the negative y-direction.
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Situation: two electrons, each with mass m and charge q, are released from positions very far from each other. With respect to a certain reference frame, electron A has initial nonzero speed v toward electron B in the positive x direction, and electron B has initial speed 3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.Question: What is the minimum separation r_min that the electrons reach?Express your answer in term of q, m, v, and k (where k=).r_min =
Answer:
[tex]r_{min}=\frac{kq^2}{5m_ev^2}[/tex]
Explanation:
The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:
[tex]E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}[/tex]
me: mass of the electron
q: charge of the electron
k: Coulomb's constant
you take into account that v2=3v1=3v and do rmin the subject of the formula:
[tex]\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}[/tex]
The minimum separation rmin of two electrons moving towards each other due to their electric repulsion can be calculated using the equation rmin = 2q2 / (mkv2), where q is the charge of the electron, m is its mass, v is the initial speed of one electron, and k is the electrostatic constant.
Explanation:When two electrons with mass m and charge q move towards each other along the x-axis, their repulsion causes them to slow down. Eventually, they reach a point where their repulsion is strong enough to push them away from each other. This point of minimum separation, denoted by rmin, can be found by equating the electrical force of repulsion to the centripetal force of motion. Since the electrical force between two point charges q1 and q2 separated by a distance r is given by F = k(q1q2)/r2, where k is the electrostatic constant, the equation for rmin is:
rmin = 2q2/ (mkv2)
where v is the initial speed of one electron towards the other.
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A trooper is moving due south along the freeway at a speed of 30.1 m/s when a red car passes the trooper. The red car moves with constant velocity of 45.4 m/s southward. At the instant the trooper's car is passed, the trooper begins to speed up at a constant rate of 2.0 m/s2. What is the maximum distance ahead of the trooper that is reached by the red car
Answer:
The correct answer to the following question will be "41.87 m".
Explanation:
The given values are:
The speed of trooper = [tex]30.1 \ m/s[/tex]
The velocity of red car = [tex]45.4 \ m/s[/tex]
Now,
A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at
[tex]t=\frac{45-30}{2.7}[/tex] (∵ [tex]time=\frac{distance}{time}[/tex])
[tex]t=5.56 \ sec[/tex]
then,
The distance covered by trooper,
[tex]t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2[/tex]
[tex]=208.33 \ m[/tex]
The distance covered by red car,
= [tex]45\times 5.56[/tex]
= [tex]250.2 \ m[/tex]
Maximum distance = [tex]250.2-208.33[/tex]
= [tex]41.87 \ m[/tex]
Final answer:
The trooper's car catches up with the red car after 7.65 seconds. The maximum distance ahead reached by the red car is 347.01 meters.
Explanation:
To find the maximum distance ahead reached by the red car, we need to calculate the time it takes for the trooper's car to catch up with the red car.
First, find the time it takes for the trooper's car to accelerate from 30.1 m/s to 45.4 m/s:
Initial velocity of the trooper's car, v₀ = 30.1 m/sFinal velocity of the trooper's car, v = 45.4 m/sAcceleration of the trooper's car, a = 2.0 m/s²Using the equation v = v₀ + at, we can solve for t:t = (v - v₀) / a = (45.4 - 30.1) / 2.0 = 7.65 seconds
Now, we can find the maximum distance ahead by multiplying the velocity of the red car by the time it took for the trooper's car to catch up:
Distance = velocity x time = 45.4 m/s x 7.65 s = 347.01 meters
A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λA, λB, and λC). She observes three distinct first-order spectral lines at the following respective angles θm (where m denotes order). θ1 = 12.9°, θ1 = 14.2°, θ1 = 15.0° (a) If the grating has 3,680 grooves per centimeter, what wavelength (in nm) describes each of these spectral lines?
Answer:
Explanation:
Given
[tex]N=3680 cm^{-1}[/tex]
therefore slit spacing [tex]d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm[/tex]
since [tex]d\sin \theta =n\lambda [/tex]
for [tex]n=1[/tex]
[tex]d\sin \theta =\lambda [/tex]
Now,at [tex]\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm[/tex]
at [tex]\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm[/tex]
at [tex]\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm[/tex]
When Brett and Will ride the carousel, Brett always selects a horse on the outside row, whereas Will prefers the row closest to the center. These rows are 19 ft 1 in.19 ft 1 in. and 11 ft 11 in.11 ft 11 in. from the center, respectively. The angular speed of the carousel is 2.72.7 revolutions per minute. What is the difference, in miles per hour, in the linear speeds of Brett and Will?
Answer:
the difference, in miles per hour, in the linear speeds of Brett and Will;
∆v = 1.38 mph
Explanation:
Given;
Angular speed w = 2.7 revolutions per minute
Converting to revolutions per hour
w = 2.7 × 60 revolutions per hour
w = 162 rev/hour
Linear speed v = angular speed × 2πr
the difference, in miles per hour, in the linear speeds of Brett and Will;
∆v = w × 2π(r1 - r2)
r1 = Brett radius in miles
r2 = Will radius in miles
r1 = 19ft 1in = (19×12 + 1) = 229 in
r1 = 229 × 1.57828283 × 10^-5 miles
r2 = 11 ft 11 in = (11×12 + 11) = 143 in
r2 = 143 × 1.57828283 × 10^-5 miles
Substituting the values;
∆v = 162 × 2π × (229-143)×1.57828283 × 10^-5 mph
∆v = 1.38 mph
Two narrow slits 0.02 mm apart are illuminated by light from a CuAr laser (λ = 633 nm) onto a screen. a) If the first fringe is 0.2 cm away from the central fringe, what is the screen distance? b) What is the angle of the first dark fringe? c) How many fringes are visible? d) What wavelength would a laser have to provide a fringe that coincides (lines up) with the third order fringe of the CuAr laser?
Answer:
a) 0.063m
b) 2.72°
c) 3151 fringes
d) 1.87*10^-6m
Explanation:
a) To find the screen distance you use the following formula:
[tex]y=\frac{m\lambda D}{d}\\\\D=\frac{dy}{m\lambda}[/tex]
D: screen distance
d: distance between slits
m: order of the fringes
λ: wavelength
By replacing the values of the parameters you obtain:
[tex]D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m[/tex]
b) The condition for dark fringes is given by:
[tex]\lambda(m+\frac{1}{2})=dsin\theta[/tex]
for the first dark fringe the angle is:
[tex]\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°[/tex]
c) the visible number of fringes is given by:
[tex]N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes[/tex]
d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:
[tex]y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m[/tex]
Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and three on the right (total mass mp=10.1 kgmp=10.1 kg ). The distance from the fulcrum to the center of the pile of rocks is rp=0.360 m.rp=0.360 m. Answer the two questions below, using three significant digits. Part A: What is the value of the torque ( ????pτp ) produced by the pile of rocks? (Enter a positive value.)
Answer:
Torque = 35.60 N.m (rounded off to 3 significant figures.
Explanation:
Given details:
The mass of the rock on the left, ms = 2.25 kg
The total mass of the rocks, mp = 10.1 kg
The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m
(a) The torque produced by the pile of rock, T = F*rp = m*g*rp
Torque = 9.8*0.360*10.1 = 35.6328
Torque = 35.60 N.m (rounded off to 3 significant figures).
g A fiber optic is made by covering a thin fiber core material with a cladding material. An optical fiber works on the principle of total internal reflection, from this we can conclude that: A) The refractive index of the core material is less than the refractive index of the cladding material. B) The refractive index of the cladding material is less than the refractive index of the core material. C) They must have the same refractive index. D) It does not matter what the refractive index is. E) The refractive index of the cladding must be exactly twice that of the core.
Answer:(b)
Explanation:
Optical fiber works on the principle of total internal reflection . Total internal reflection occurs when the following two conditions are met
(1) When the angle of incidence of light is greater than the critical angle and,
(2) When a ray of light travels from a optically denser medium to optically rarer medium .
We know that light bend away from normal when it travels from denser to rarer medium. If we we choose such an angle that the refraction angle is [tex]90^{\circ}[/tex] then the incident angle is called critical angle.
So, the refractive index of cladding must be must be less than the refractive index of core material.
Rank the following meteorites in terms of their age, from youngest to oldest:1) a meteorite containing 500 238U isotopes and 100 206Pb isotopes.2) a meteorite containing 400 238U isotopes and 100 206Pb isotopes.3) a meteorite containing 400,000 238U isotopes and 400,000 206Pb isotopes.4) a meteorite containing 4,000 238U isotopes and 200 206Pb isotopes.5) a meteorite containing 3,000 238U isotopes and 1,000 206Pb isotopes.
Answer:
Explanation:
When 238U which is radioactive turns into 206Pb , it becomes stable and no further disintegration is done . Hence in the initial period ratio of 238U undecayed and 206Pb formed will be very high because no of atoms of 238U in the beginning will be very high. Gradually number of 238U undecayed will go down and number of 206Pb formed will go up . In this way the ratio of 238U and 206Pb in the mixture will gradually reduce to be equal to one or even less than one .
In the given option we shall calculate their raio
1 ) ratio of 238U and 206Pb = 5
2 ) ratio of 238U and 206Pb = 4
3 )ratio of 238U and 206Pb = 1
4 ) ratio of 238U and 206Pb = 20
5 )ratio of 238U and 206Pb = 3
lowest ratio is 1 , hence this sample will be oldest.
Ranking from youngest to oldest
4 , 1 , 2 , 5 , 3 .
1. Laser beam with wavelength 632.8 nm is aimed perpendicularly at opaque screen with two identical slits on it, positioned horizontally, and close enough so that both of them fall in the beam cross section. By shifting vertically the screen, each slits can be illuminated independently, allowing the other to be disregarded. On an observation screen positioned 1m further from the opaque screen, the diffraction patterns from the independent slit illumination were found identical, with minima 6.5 mm apart. The interference pattern maxima from simultaneous illumination of both slits were 0.53 mm apart. What are the double slit characteristics (slit width, slit separation)
To find the double-slit characteristics of width and separation, one must use the data from a single-slit diffraction pattern to calculate the slit width and then the interference pattern from both slits to find the slit separation.
The student's question is about the characteristics of a double-slit diffraction pattern produced when a laser beam with a specific wavelength is projected through two slits. To deduce the slit width and separation, we use the information provided about the diffraction and interference patterns observed.
Firstly, we use the minima separation from the single slit diffraction to find the slit width using the diffraction formula d sin(θ) = mλ, where d is the slit width, m is an integer denoting the order of the minima, λ is the wavelength, and θ is the angle of the minima. Since the screen is 1 meter away and the minima are 6.5 mm apart, we can approximate sin(θ) as the ratio of minima separation to the distance to the screen to solve for the slit width d.
Next, we apply the formula for double-slit interference, d sin(θ) = nλ (where d is now the slit separation, n is an integer denoting the order of the maxima) to the observed interference maxima separation to calculate the slit separation.
Slit separation: [tex]\(97.23 \times 10^{-6}\)[/tex] m. Slit width: [tex]\(1194.34 \times 10^{-6}\)[/tex] m. Calculated using interference formula and provided data.
To solve this problem, we can use the double-slit interference formula:
[tex]\[ \Delta y = \frac{\lambda L}{d} \][/tex]
where:
- [tex]\( \Delta y \)[/tex] is the distance between adjacent maxima or minima on the observation screen,
- [tex]\( \lambda \)[/tex] is the wavelength of the laser beam,
- [tex]\( L \)[/tex] is the distance between the screen and the slits, and
- [tex]\( d \)[/tex] is the slit separation.
Given:
- Wavelength [tex]\( \lambda = 632.8 \)[/tex] nm = [tex]\( 632.8 \times 10^{-9} \)[/tex] m,
- Distance between the screen and the slits [tex]\( L = 1 \)[/tex] m = 1 m,
- Distance between adjacent minima [tex]\( \Delta y_{\text{min}} = 6.5 \)[/tex] mm = [tex]\( 6.5 \times 10^{-3} \)[/tex] m, and
- Distance between adjacent maxima [tex]\( \Delta y_{\text{max}} = 0.53 \)[/tex] mm = [tex]\( 0.53 \times 10^{-3} \)[/tex] m.
First, let's find the slit separation [tex]\( d \)[/tex] using the minima data:
[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} \][/tex]
Then, let's find the slit width [tex]\( w \)[/tex] using the maxima data:
[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} \][/tex]
Let's calculate these values.
First, let's calculate the slit separation [tex]\(d\)[/tex] using the minima data:
[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{6.5 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ d \approx \frac{(632.8)(1)}{6.5} \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ d \approx 97.23 \times 10^{-6} \, \text{m} \][/tex]
Now, let's calculate the slit width [tex]\(w\)[/tex] using the maxima data:
[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{0.53 \times 10^{-3} \, \text{m}} \][/tex]
[tex]\[ w \approx \frac{(632.8)(1)}{0.53} \times 10^{-6} \, \text{m} \][/tex]
[tex]\[ w \approx 1194.34 \times 10^{-6} \, \text{m} \][/tex]
So, the slit separation [tex]\(d\)[/tex] is approximately [tex]\(97.23 \times 10^{-6}\)[/tex] m, and the slit width [tex]\(w\)[/tex] is approximately [tex]\(1194.34 \times 10^{-6}\)[/tex] m.
What is one way to increase the amplitude if a wave in a medium?
A. By applying a vibration at the natural frequency of the medium
B. By increasing the velocity of the wave source
C. By moving the source of the wave away from the receptor or observer
D. By increasing the frequency of the wave source
Answer: c
Explanation:
1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the collision. Calculate the resulting speed for an inelastic collision (when they stick together). 2. A small object (m = 200 g) collides elastically with a larger object (m = 1000 g), which was at rest before the collision. The incoming speed of the smaller object was 1.0 m/s. The speed of the larger object after the collision is 0.33 m/s. Calculate the resulting speed and determine the direction for the smaller object after the collision when it rebounds. (Watch out for the directions of the motions and use respective signs for the velocities and momentums.)
Answer:
Explanation:
1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ = [tex]\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}[/tex]
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
= [tex]\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}[/tex]
= - .66 m /s
Since the sign is negative so it will be in opposite direction .
1. The resultant speed for an inelastic collision is =.05 m /s
2. When The direction for the smaller object the collision when it rebounds is = -.66 m /s
What is Momentum of Law?
Conservation of momentum is refer to a major law of physics that states that the momentum of a system is constant if no external forces are acting on the system. when It is embodied in Newton’s First Law or The Law of Inertia.
1. We are involving the conservation of momentum law to solve the problem.
mv is = ( M +m) V, m, and M are masses of the small and large objects, v is the velocity of a small object before the collision and also V is the velocity of both the objects together after the collision.
Then .5 x .2 = (1.5 + .5)V
Therefore, V = .05 m /s
2. We shall use the formula for the velocity of the object behind the elastic collision as follows
After that, v₁ = (m₁-m₂)/(m₁+m₂)u1 + 2m₂u₂/(m₁+m₂)
m₁ and m₂ are masses of the first and second objects u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
Now, = (200-1000)/(1000+200) × 1 + 2×1000×0/(1000+200)
Thus, = - .66 m /s
When the sign is negative so it will be in opposite direction.
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The electric field at the center of a ring of charge is zero. At very large distances from the center of the ring along the ring's axis the electric field goes to zero. Find the distance from the center of the ring along the axis (perpendicular to the plane containing the ring) at which the magnitude of the electric field is a maximum. The radius of the ring is 6.58 cm and the total charge on the ring is 8.87E-6 C.
Answer:
z = 1.16m
Explanation:
The electric field in a point of the axis of a charged ring, and perpendicular to the plane of the ring is given by:
[tex]E_z=\frac{Qz}{(z^2+r^2)^{\frac{3}{2}}}[/tex]
z: distance to the plane of the ring
r: radius of the ring
Q: charge of the ring
you have that:
E_{z->0} = 0
E_{z->∞} = 0
To find the value of z that maximizes E you use the derivative respect to z, and equals it to zero:
[tex]\frac{dE_z}{dz}=Q[\frac{1}{(z^2+r^2)^{3/2}}+z(-\frac{3}{2})\frac{1}{(z^2+r^2)^{5/2}}(2z)]=0\\\\(z^2+r^2)^{5/2}=3z^2(z^2+r^2)^{3/2}\\\\(z^2+r^2)^2=3z^4\\\\z^4+2z^2r^2+r^4=3z^4\\\\2z^4-2z^2r^2-r^4=0\\\\z^2_{1,2}=\frac{-(-2)+-\sqrt{4-4(2)(-1)}}{2(2)}=\frac{2\pm 3.464}{4}\\\\[/tex]
you take the positive value:
[tex]z^2=\frac{2+3.464}{4}=1.366\\\\z=1.16m[/tex]
hence, the distance in which the magnitude if the electric field is maximum is 1.16m
Water is completely filling black metallic vessel having cubic form and thin walls. The mass of water is 1 kg and initial temperature 50 C. This cubic vessel is placed into black cavity whose walls are kept at the constant temperature of 0 C. Find time t it takes for the water to cool down to 10 C. Water specific heat is 4200 (J/kg K-1 .) After you find an answer, estimate whether this radiation mechanism would be responsible for the real-life water cooling and what other mechanisms might be in play.
Answer:
Check the attached image
Explanation:
To solve the problem for time you will have to use the formula for time, t = d/s which means time equals distance divided by speed.
Kindly check the attached image below for the step by step explanation to the question.
A 17.0 m long, thin, uniform steel beam slides south at a speed of 28.0 m/s. The length of the beam maintains an east-west orientation while sliding. The vertical component of the Earth's magnetic field at this location has a magnitude of 30.0 µT. What is the magnitude of the induced emf between the ends of the beam (in mV)?
Answer:14.2 mV
Explanation:
Given
Length of steel beam [tex]L=17\ m[/tex]
Magnetic field [tex]B=30\ \muT[/tex]
speed of beam [tex]v=28\ m/s[/tex]
Now consider as beam slides it encloses an area of rectangle of
width [tex]w=17\ m and length L'=v\times t [/tex]
Area [tex]A=17v\cdot t[/tex]
and Induced EMF is [tex]V=-\frac{d(BA)}{dt}[/tex]
[tex]V=-B\frac{d(17vt)}{dt}[/tex]
[tex]V=-17Bv[/tex]
[tex]V=-17\times 30\times 10^{-6}\times 28[/tex]
[tex]V=-14.2\ mV[/tex]
Magnitude of EMF[tex]=14.2\ mV[/tex]
Consider a steel guitar string of initial length L=1.00L=1.00 meter and cross-sectional area A=0.500A=0.500 square millimeters. The Young's modulus of the steel is Y=2.0×1011Y=2.0×1011 pascals. How far ( ΔLΔLDelta L) would such a string stretch under a tension of 1500 newtons?
Answer:
The extent to which it would stretch is [tex]\Delta L = 0.015 \ m[/tex]
Explanation:
From the question we are told that
The initial length is [tex]L = 1.00m[/tex]
The area is [tex]A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2[/tex]
The Young modulus of the steel is [tex]Y = 2.0*10^{11} Pa[/tex]
The tension is [tex]T =1500 N[/tex]
The Young modulus is mathematically represented as
[tex]Y = \frac{\sigma}{e}[/tex]
Where [tex]\sigma[/tex] is the stress which is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
Substituting values
[tex]\sigma = \frac{1500}{0.500*10^{-6}}[/tex]
[tex]\sigma = 3.0*10^9 N/m^2[/tex]
And e is the strain which is mathematically represented as
[tex]e = \frac{\Delta L}{L }[/tex]
Where [tex]\Delta L[/tex] The extension of the steel string
Substituting these into the equation above
[tex]Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]
Substituting values
[tex]2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]
[tex]\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}[/tex]
[tex]\Delta L = 0.015 \ m[/tex]
Un ciclista recorre una pista recta de ida y vuelta, en este recorrido:
A) La distancia total recorrida es cero.
B) La aceleración es distinta de cero.
C) El tiempo total es cero.
D) La rapidez es cero.
E) El desplazamiento total es cero.
Answer:
a) falso
b) verdadero
c) falso
d) falso
e) verdadero
Explanation:
A) FALSO: la distancia recorrida es independiente del punto de partida y del punto de llegada
B) VERDADERO: para que el ciclista recorra la pista de ida y vuelta es necesario que su velocidad cambie, por lo tanto la aceleración es diferente de cero.
C) FALSO
D) FALSO: en todo momento hay una distancia recorrida en un tiempo en específico, es decir, una rapidez.
E) VERDADERO: el despazamiento sí depende de la distancia entre el punto de llegada y el punto de salida.
Migratory birds are able to use the earth's magnetic field to navigate even when clouds and darkness prevent them from having visual references for their flight. The range of sensitivity of such birds extends to magnetic fields as small as about a third of the earth's natural field. If such a bird is flying past a power line which carries 105 amps, and if we assume that the minimum field detectable at 60 Hz is the same as the minimum field detected at DC, at what distance could the bird detect the presence of the power line? Why will the real answer be much smaller? (think about how power lines are arranged)
Answer:
The receptors that sense the Earth's magnetic field are probably located in the birds' eyes. Now, researchers at Lund University have studied different proteins in the eyes of zebra finches and discovered that one of them differs from the others: only the Cry4 protein maintains a constant level throughout the day and in different lighting conditions
Explanation:
Final answer:
The bird will be able to detect the power line at a much smaller distance than calculated due to the concentration of the magnetic field close to the wire.
Explanation:
The distance at which a migratory bird could detect the presence of a power line can be calculated using the equation for the magnetic field produced by a long straight wire. However, since the bird's sensitivity to magnetic fields is much smaller than the field produced by the power line, the real answer will be much smaller than the calculated distance. This is because power lines are typically arranged in such a way that the magnetic field they produce is concentrated close to the wire, diminishing rapidly with distance.
Two stereo speakers mounted 4.52 m apart on a wall emit identical in-phase sound waves. You are standing at the opposite wall of the room at a point directly between the two speakers. You walk 2.11 m parallel to the wall, to a location where you first notice that the sound intensity drops to zero.
If the wall along which you are walking is 10.7 m from the wall with the speakers, what is the wavelength of the sound waves?
a) 2.1 m
b) 2.9 m
c) 0.9 m
d) 1.7 m
The wavelength of the sound waves is equal to 1.714 m. Therefore, option (d) is correct.
What are sound waves?Sound waves can transmit through liquids, and gases, as well as plasma as longitudinal waves, called compression waves. It requires a medium to travel so it can be traveled through solids in the form of both longitudinal waves and transverse waves.
Given, the distance between two speakers, d₁ = 4.52 m
The distance between two walls, d₂ = 10.2 m
The expression to calculate wavelength: pd =(n + 0.5 λ)
As there is no slit, n =0 therefore, pd = 0.5 λ ...........(1)
Given, the triangle-1, the value of s₁ = 10.7 m
The value of s₂ = (4.52/2) - 2.11 = 0.15 m
The value of s₃ is equal to: [tex]s_3 =\sqrt{(10.7)^2+(0.15)^2}[/tex]
s₃ = 10.7 m
Given, the triangle-2, the value of a₁ = 10.7 m
The value of a₂ = (4.52/2) + 2.11 = 4.37 m
The value of s₃ is equal to: [tex]s_3 =\sqrt{(10.7)^2+(4.37)^2}[/tex]
a₃ = 11.557 m
The constructive interference, pd = a₃ - s₃ = 11.557 - 10.7 = 0.857 m
From the equation (1) we can determine wavelength:
pd = 0.5 λ
0.857 = 0.5 λ
λ = 1.714 m
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2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initial temperature at 1336K. The solid and liquid can only exchange heat with each other. What kind of analysis do you need to perform in order to determine whether, once thermal equilibrium is reached, the mixture will be entirely solid or in a mixed solid/liquid phase?
Answer:
Explanation:
The specific heat of gold is 129 J/kgC
It's melting point is 1336 K
It's Heat of fusion is 63000 J/kg
Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,
The first is E1 = 63000 J/kg x 1.5 = 94500 J
the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J
Therefore, all solid is not correct. You will have a mixture of solid and liquid.
For more detail, the difference between E1 and E2 is 7812 J, and that will melt
7812/63000 = 0.124 kg of the solid gold
The question involves a thermodynamic analysis using heat transfer and phase change principles to determine if a mixture of solid and liquid gold will become entirely solid or remain mixed upon reaching thermal equilibrium.
Explanation:To determine whether the mixture of 2.0 kg of solid gold at 1000K and 1.5 kg of liquid gold at 1336K will be entirely solid or in a mixed solid/liquid phase upon reaching thermal equilibrium, a thermodynamic analysis involving the principles of heat transfer and phase change is required. This analysis will utilize the concept of conservation of energy, accounting for the specific heats of solid and liquid gold as well as the latent heat of fusion if a phase change occurs. The calculations will involve setting the heat lost by the liquid gold equal to the heat gained by the solid gold until both reach the same temperature. If the final temperature is below the melting point of gold without all the solid gold melting, the result will be a mixture. If all the solid melts before reaching the equilibrium temperature, the final state will be all liquid. It's critical to remember that the process must comply with the melting point of gold, which is 1064K.
A large turntable with radius 6.00 m rotates about a fixed vertical axis, making one revolution in 8.00 s. The moment of inertia of the turntable about this axis is 1200 kg⋅m². You stand, barefooted, at the rim of the turntable and very slowly walk toward the center, along a radial line painted on the surface of the turntable. Your mass is 73.0 kg. Since the radius of the turntable is large, it is a good approximation to treat yourself as a point mass. Assume that you can maintain your balance by adjusting the positions of your feet. You find that you can reach a point 3.00 m from the center of the turntable before your feet begin to slip.
What is the coefficient of static friction between the bottoms of your feet and the surface of the turntable?
Answer:
0.8024
Explanation:
From the given question; we can say that the angular momentum of the system is conserved if the net torque is is zero.
So; [tex]I_o \omega _o = I_2 \omega_2[/tex]
At the closest distance ; the friction is :
[tex]f_s = \mu_s (mg)[/tex]
According to Newton's Law:
F = ma
F = mrω²
From conservation of momentum:
[tex]I_o \omega _o = I_2 \omega_2[/tex]
[tex]\omega_2= \frac { I_o \omega _o}{ I_2 }[/tex]
[tex]\omega_2=( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })* \omega_o[/tex]
However ; since the static friction is producing the centripetal force :
[tex]\mu_s (mg) = mr \omega_2^2[/tex]
[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]
The coefficient of static friction between the bottoms of your feet and the surface of the turntable can now be calculated by using the formula :
[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]
= [tex]( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })^2 * \frac{ \omega^2_o *r }{g}[/tex]
= [tex][\frac{1200+(73(6)^2)}{((1200)+(73)(3)^2)} ]^2*[\frac{(\frac{\pi}{4})^2(3)} {9.8}][/tex]
= 0.8024
The friction force between the bottom of the feet and the surface of the
turntable balance the centrifugal force due to rotation.
The coefficient static friction is approximately 0.802Reasons:
Radius of the turntable, r₀ = 6.00 m
The period of rotation of the turntable, T₀ = 8.00s
Moment of inertia of the turntable, [tex]I_t[/tex] = 1,200 kg·m²
Mass of the person, m = 73.0 kg
The point at which the feet starts to slip, r₁ = 3.00 m
Required:
The friction between the bottom of the feet and the surface of the turntable.
Solution:
The moment of inertia of the person and the turntable combined, J, can be
found by considering the person as a point mass.
Therefore;
At the rim, J₀ = [tex]I_t[/tex] + m·r² = 1,200 + 73×6² = 3828
J₀ = 3,828 kg·m²
At 3.00 m from the center, J₁ = [tex]I_t[/tex] + m·r² = 1,200 + 73×3² = 1,857
J₁ = 1,857 kg·m²
Angular momentum, L = J·ω
Whereby the angular momentum is conserved, we have;
L₀ = L₁
J₀·ω₀ = J₁·ω₁
[tex]\displaystyle \omega_1 = \mathbf{ \frac{J_0 \cdot \omega_0}{J_1}}[/tex]
Which gives;
To remain at equilibrium, the friction force, [tex]F_f[/tex] = The centrifugal force, [tex]F_c[/tex]
[tex]F_f =\mu_s \cdot W = \mu_s \cdot m \cdot g[/tex][tex]\displaystyle F_c = \frac{m \cdot v^2}{r} = \frac{m \cdot (\omega \cdot r)^2}{r} = m \cdot r \cdot \omega^2[/tex]
Where;
[tex]\mu_s[/tex] = The coefficient of static friction
g = Acceleration due to gravity which is approximately 9.81 m/s²
r = The radius of rotation
ω = The angular speed
[tex]F_f[/tex] = [tex]F_c[/tex]
Therefore;
[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{m \cdot r \cdot \omega^2}[/tex]Therefore, at 3.00 m from the center of the turntable, we have;
[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{ m \cdot r_1 \cdot \omega_1^2}[/tex]
[tex]\displaystyle\mu_s \cdot g = r_1 \cdot \omega_1^2[/tex]
[tex]\displaystyle\mu_s = \frac{r_1 \cdot \omega_1^2}{g} = \mathbf{\frac{r_1 \cdot \left(\displaystyle \frac{J_0 \cdot \omega_0}{J_1}\right)^2}{g}}[/tex]
Which gives;
[tex]\displaystyle\mu_s =\frac{3 \times \left(\displaystyle \frac{3,828 \times \frac{2 \cdot \pi}{8} }{1,857}\right)^2}{9.81} \approx \mathbf{0.802}[/tex]
The coefficient of static friction between the bottom of the feet and the surface of the turntable, at 3.00 m from the center, [tex]\mu_s[/tex] ≈ 0.802.
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A cyclist is moving up a slope that is at an angle of 19 to the horizontal. The mass of the cyclist and the bicycle is 85 kg. What is the component of the weight of the cyclist and bicycle parallel to the slope and what is the normal reaction force on the bicycle from the slope?
Answer:
Explanation:
Given
Slope of inclination [tex]\theta =19^{\circ}[/tex]
Mass of cyclist and bicycle is [tex]m=85\ kg[/tex]
When cyclist is going up then there is two components of its weight , one is parallel to inclination and other is perpendicular to inclination
Weight of person is mg
it can be resolved in [tex]mg\cos \theta[/tex] and [tex]mg\sin \theta [/tex]
[tex]mg\cos \theta[/tex] is perpendicular to the inclination
and [tex]mg\sin \theta[/tex] is parallel to inclination as shown in diagram
Parallel component [tex]=mg\sin \theta=85\times 9.8\times \sin 19=271.2\ N[/tex]
Normal reaction [tex]N=mg\cos \theta =85\times 9.8\times \cos 19=787.61\ N[/tex]
The normal reaction on the bicycle by the inclined slope is 787.62 N.
The parallel on the bicycle along the inclined slope is 271.2 N.
The given parameters;
angle of the slope, = 19mass of the bicycle, m = 85 kgThe normal reaction on the bicycle by the inclined slope is calculated as; follows;
[tex]F_n = W cos \theta\\\\F_n = mg \ cos \theta \\\\F_n = 85 \times 9.8 \times cos \ (19)\\\\F_n = 787.62 \ N[/tex]
The parallel on the bicycle along the inclined slope is calculated as
[tex]F_x = Wsin\theta\\\\F_x = mg \ sin \ \theta \\\\F_x = 85 \times 9.8 \times sin(19) \\\\F_ x = 271 .2 \ N[/tex]
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You've recently read about a chemical laser that generates a 20.0-cm-diameter, 30.0 MW laser beam. One day, after physics class, you start to wonder if you could use the radiation pressure from this laser beam to launch small payloads into orbit. To see if this might be feasible, you do a quick calculation of the acceleration of a 20.0-cm-diameter, 99.0 kg, perfectly absorbing block.
What speed would such a block have if pushed horizontally 100 m along a frictionless track by such a laser?
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the distance of point charge on the line joining the charge B. Where the resultant electric field is zero
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
An undamped 2.47 kg2.47 kg horizontal spring oscillator has a spring constant of 32.8 N/m.32.8 N/m. While oscillating, it is found to have a speed of 2.30 m/s2.30 m/s as it passes through its equilibrium position. What is its amplitude AA of oscillation? A=A= mm What is the oscillator's total mechanical energy EtotEtot as it passes through a position that is 0.6520.652 of the amplitude away from the equilibrium position? Etot=Etot= J
Answer:
0.631 m
6.53315 J
Explanation:
m = Mass = 2.47 kg
v = Velocity = 2.30 m/s
k = Spring constant = 32.8 N/m
A = Amplitude
In this system the energy is conserved
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{2.47\times 2.3^2}{32.8}}\\\Rightarrow A=0.631\ m[/tex]
The amplitude is 0.631 m
Mechanical energy is given by
[tex]E=\dfrac{1}{2}mv^2\\\Rightarrow E=\dfrac{1}{2}2.47\times 2.3^2\\\Rightarrow E=6.53315\ J[/tex]
The mechanical energy is 6.53315 J
A submarine is 2.84 102 m horizontally from shore and 1.00 102 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 2.34 102 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. The goal is to find the height of the target above sea level.
Answer:
468 m
Explanation:
So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.
The length CE is length AE - length AC = 284 - 234 = 50 m
We can calculate the angle ECD:
[tex]tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2[/tex]
[tex]\hat{ECD} = tan^{-1} 2 = 63.43^o[/tex]
This is also the angle ACB, so we can find the length AB:
[tex]tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}[/tex]
[tex]2 = \frac{AB}{234}[/tex]
[tex]AB = 2*234 = 468 m[/tex]
So the height of the building is 468m