Answer:
Explanation:
Magnetic field due to moving charge =
[tex]B=\frac{\mu_0}{4\pi} \times \frac{qv}{r^2}[/tex]
q is charge moving with velocity v and r is distance from point at which field is calculated .
For alpha particle
[tex]B_1=\frac{\mu_0}{4\pi} \times \frac{2\times 1.6\times10^{-19}\times 2.8\times 10^5}{(8.25\times 10^{-9})^2}[/tex]
= 0.1316 x 10⁻³ T
For electron
[tex]B_2=\frac{\mu_0}{4\pi} \times \frac{ 1.6\times10^{-19}\times 2.8\times 10^5}{(8.25\times 10^{-9})^2}[/tex]
= .0658 x 10⁻³ T .
Both these magnetic field will be same in direction because direction of equivalent current is same for both the particles .
Hence
Total magnetic field
= B₁ + B₂ = .1974 x 10⁻³
= 1.974 x 10⁻⁴ T .
The total magnetic field produced by these charges at the given point is 1.98 x 10⁻⁴ T.
The given parameters;
charge of the alpha-particle, q = 2espeed of the charges, v = 2.8 x 10⁵ m/sseparation distance, r = 8.25 nmThe magnetic field produced by each charge is calculated using Biot-Savart law;
[tex]B = \frac{\mu _o q}{4\pi } \times \frac{v}{r^2} \\\\B_1 = \frac{(4\pi \times 10^{-7} ) \times (1.602 \times 10^{-19})}{4\pi } \times \frac{2.8 \times 10^5}{(8.25 \times 10^{-9} )^2}\\\\B_1 = 6.6 \times 10^{-5} \ T[/tex]
[tex]B_2 = \frac{(4\pi \times 10^{-7} ) \times (2\times 1.602 \times 10^{-19})}{4\pi } \times \frac{2.8 \times 10^5}{(8.25 \times 10^{-9} )^2}\\\\B_2 = 1.32 \times 10^{-4} \ T[/tex]
The total magnetic field produced by these charges at the given point P;
[tex]B_T = B_1 + B_2\\\\B_T = 6.6\times 10^{-5} \ + \ 1.32 \times 10^{-4}\\\\B_T = 1.98 \times 10^{-4} \ T[/tex]
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The low-frequency speaker of a stereo set has a surface area of 0.07 m2 and produces 1.63 W of acoustical power. What is the intensity at the speaker (in W/m2)?
Answer:
I = 81.5 W/m^2
Explanation:
In order to calculate the intensity of the sound at the speaker, you use the following formula:
[tex]I=\frac{P}{A}[/tex] (1)
P: power of the speaker's sound = 1.63W
A: surface area of the stereo set = 0.07m^2
You assume that the intensity of the sound at the speaker depends only of the surface area of the stereo set. Furthermore, you consider that the wave front of the sound is approximately plane.
You replace the values of the parameters in the equation (1):
[tex]I=\frac{1.63W}{0.02m^2}=81.5\frac{W}{m^2}[/tex]
The intensity of the speaker's sound at the speaker is 81.5 W/m^2
A 60-watt light bulb carries a current of 0.5 ampere. The total charge passing through it in one hour is:
Answer:
Total charge = 1800C
Explanation:
Q= IT
I = currentt = timeQ = chargeAn electric motor drawing 10 amps at 110 V in steady state produces shaft power at 9.7 Nm and 1000 RPM. For a first Law analysis and considering the motor as the control volume How much heat will be produced from the motor (in Watts)
Answer:
The heat rate produced from the motor is 84.216 watts.
Explanation:
The electric motor receives power from electric current and releases power in the form of mechanical energy (torque) and waste heat and can be considered an stable-state system. The model based on the First Law of Thermodynamics for the electric motor is:
[tex]\dot W_{e} - \dot W_{T} -\dot Q = 0[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer from the electric motor, measured in watts.
[tex]\dot W_{e}[/tex] - Electric power, measured in watts.
[tex]\dot W_{T}[/tex] - Mechanical power, measured in watts.
The heat transfer rate can be calculated in terms of electric and mechanic powers, that is:
[tex]\dot Q = \dot W_{e} - \dot W_{T}[/tex]
The electric and mechanic powers are represented by the following expressions:
[tex]\dot W_{e} = i \cdot V[/tex]
[tex]\dot W_{T} = T \cdot \omega[/tex]
Where:
[tex]i[/tex] - Current, measured in amperes.
[tex]V[/tex] - Steady-state voltage, measured in volts.
[tex]T[/tex] - Torque, measured in newton-meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now, the previous expression for heat transfer rate is expanded:
[tex]\dot Q = i \cdot V - T \cdot \omega[/tex]
The angular speed, measured in radians per second, can be obtained by using the following expression:
[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]
Where:
[tex]\dot n[/tex] - Rotational rate of change, measured in revolutions per minute.
If [tex]\dot n = 1000\,rpm[/tex], then:
[tex]\omega = \left(\frac{\pi}{30} \right)\cdot (1000\,rpm)[/tex]
[tex]\omega \approx 104.720\,\frac{rad}{s}[/tex]
Given that [tex]i = 10\,A[/tex], [tex]V = 110\,V[/tex], [tex]T = 9.7\,N\cdot m[/tex] and [tex]\omega \approx 104.720\,\frac{rad}{s}[/tex], the heat transfer rate from the electric motor is:
[tex]\dot Q = (10\,A)\cdot (110\,V) -(9.7\,N\cdot m)\cdot \left(104.720\,\frac{rad}{s} \right)[/tex]
[tex]\dot Q = 84.216\,W[/tex]
The heat rate produced from the motor is 84.216 watts.
Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
W = 12.96 J
A simple random sample is a sample drawn in such a way that each member of the population has some chance for being included in the sample every tenth element of an arranged population is included each member of the population has equal chance for being included in the sample each member of the population has 0.10 chance for being included in the sample:__________.
A simple random sample is a sample drawn in such a way that each member of the population has equal chance for being included in the sample.
The mass percent of hydrogen in CH₄O is 12.5%.
What is the mass percent?Mass percent is the mass of the element divided by the mass of the compound or solute.
Step 1: Calculate the mass of the compound.
mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu
Step 2: Calculate the mass of hydrogen in the compound.
mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu
Step 3: Calculate the mass percent of hydrogen in the compound.
%H = (mH in mCH₄O / mCH₄O) × 100%
%H = 4.00 amu / 32.01 amu × 100% = 12.5%
The mass percent of hydrogen in CH₄O is 12.5%.
CO2 = 1.580 grams H2O = 0.592 grams Lookup the molar mass of each element in the compound Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Calculate the molar mass of CH4O by adding the total masses of each element used. 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 Now calculate how many moles of CH4O you have by dividing by the molar mass. m = 1.15 g / 32.04146 g/mole = 0.035891 mole Now figure out how many moles of carbon and hydrogen you have. Carbon = 0.035891 moles Hydrogen = 0.035891 moles *
Therefore, The mass percent of hydrogen in CH₄O is 12.5%.
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1. A base-ball of mass 0.3kg approaches the bat at a speed of 30 miles/hour and when the ball hits the bat for 0.5 s, it started to move away from the bat at a speed of 60 miles/hour. Find the impulse
Answer:
I = 27kg.mi/h
Explanation:
In order to calculate the impulse of the ball, you use the following formula:
[tex]I=m\Delta v[/tex] [tex]=m(v-v_o)[/tex] (1)
m: mass of the ball = 0.3kg
v: speed of the ball after the bat hit it = 60mi/h
vo: speed of the ball before the bat hit it = 30mi/h
You replace the values of all parameters in the equation (1):
[tex]I=(0.3kg)(60mi/h-(-30mi/h))=27kg\frac{mi}{h}[/tex]
where the minus sign of the initial velocity means that the motion of the ball is opposite to the final direction of such a motion.
The imulpse of the ball is 27 kg.miles/hour
A student in her physics lab measures the standing-wave modes of a tube. The lowest frequency that makes a resonance is 30 Hz. As the frequency is increased, the next resonance is at 90 Hz.
What will be the next resonance after this?
Answer:
The next resonance will be 150 Hz.
Explanation:
The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.
The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.
Frecuency of the standing sound wave modes in a open-closed tube is:
fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)
The next resonance is at 90 Hz. This means that it occurs when n = 3:
f₃=3*30 Hz= 90 Hz
This means that the next resonance occurs when n = 5:
f₅=5*30 Hz= 150 Hz
The next resonance will be 150 Hz.
4. How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other
Answer:
Explanation:
Pls see diagram in attached file
A brick is dropped with zero initial speed from the roof of a building and strikes the ground in 1.90 s. How tall is the building?
Answer:
17.69 m
Explanation:
The time it takes the brick to strike the ground is 1.90 seconds.
We can apply one of Newton's equation of linear motion to find the height of the building:
[tex]s = ut + 0.5gt^2[/tex]
where s = distance (in this case height)
u = initial velocity = 0 m/s
t = time = 1.90 s
g = acceleration due to gravity = 9.8 m/s^2
Therefore:
s = (0 * 1.9) + (0.5 * 9.8 * 1.9 * 1.9)
s = 0 + 17.68
s = 17.69 m
The height of the building is 17.69 m.
A 75kg passenger at the bottom of a roller coaster loop that has a radius of 20m. If the roller coaster car is moving 10m/s, what is the apparent weight of the passenger? g
Answer:
The apparent weight of the passenger is 360 N
Explanation:
Given;
The mass of the passenger, m = 75 kg
radius of the loop, r = 20 m
velocity of the roller coaster, v = 10 m/s
Centripetal force acting on this passenger is given as;
[tex]F = \frac{mv^2}{r}[/tex]
where;
F is the centripetal force acting on the passenger
m is the mass of the passenger
v is the velocity of the passenger
r is the radius of the track
[tex]F = \frac{mv^2}{r} \\\\F = \frac{75*10^2}{20} \\\\F = 375 \ N[/tex]
Real weight of the passenger,
W = mg
where;
g is acceleration due to gravity
W = 75 x 9.8
W = 735 N
Apparent weight of the passenger = Real weight - Centripetal force
Apparent weight of the passenger = 735 N - 375 N
Apparent weight of the passenger = 360 N
Therefore, the apparent weight of the passenger is 360 N
A diver wants to jump from a board, the initial height is 10 meters and he wants to reach a horizontal distance of 2 meters. What minimum speed must he have when jumping from the board to achieve his goal?
Answer:
1.4 m/s
Explanation:
The minimum speed will be when the diver's initial velocity is horizontal.
First, find the time it takes for the diver to fall 10 meters.
Given:
Δy = 10 m
v₀ᵧ = 0 m/s
aᵧ = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
10 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.43 s
Now find the initial horizontal velocity.
v = (2 m) / (1.43 s)
v = 1.4 m/s
The speed of light in a material is 1.74 x 108 m's. What is the index of refraction of this material?
Answer:
The index of refraction of this material is 1.7241
Explanation:
Recall that the index of refraction (n) of a medium is defined as the quotient between the speed of light in vacuum divided the speed of light in the medium whose index of refraction is being calculated. In mathematical terms:
[tex]n=\frac{c}{v}[/tex]
Therefore, in our case, since we know the speed of light in the medium ([tex]v=1.74\,\,10^8\,\,m/s[/tex]) and the speed of light in vacuum ([tex]c=3\,\,10^8\,\,m/s[/tex]), we can estimate the index of refraction of the medium:
[tex]n=\frac{c}{v} \\n=\frac{3\,\,10^8}{1.74\,\,10^8} \\n=1.7241[/tex]
To understand the behavior of the electric field at the surface of a conductor, and its relationship to surface charge on the conductor. A conductor is placed in an external electrostatic field. The external field is uniform before the conductor is placed within it. The conductor is completely isolated from any source of current or charge.
1) Which of the following describes the electricfield inside this conductor?
A) It is in the same direction as the original external field.
B) It is in the opposite direction from that of the original external field.
C) It has a direction determined entirely by the charge on its surface.
D) It is always zero.
2) The charge density inside the conductor is:______.
a) 0.
b) non-zero;but uniform.
c) non-zero;non-uniform.
d) infinite.
3) Assume that at some point just outside the surface of the conductor, the electric field has magnitudeE and is directed toward the surface of the conductor. What is the charge density eta on the surface of the conductor at that point?
Answer:
1) Option D is correct.
The electric field inside a conductor is always zero.
2) Option A is correct.
The charge density inside the conductor is 0.
3) Charge density on the surface of the conductor at that point = η = -E ε₀
Explanation:
1) The electric field is zero inside a conductor. Any excess charge resides entirely on the surface or surfaces of a conductor.
Assuming the net electric field wasn't zero, current would flow inside the conductor and this would build up charges on the exterior of the conductor. These charges would oppose the field, ultimately (in a few nanoseconds for a metal) canceling the field to zero.
2) Since there are no charges inside a conductor (they all reside on the surface), it is logical that the charge density inside the conductor is also 0.
3) Surface Charge density = η = (q/A)
But electric field is given as
E = (-q/2πε₀r²)
q = -E (2πε₀r²)
η = (q/A) = -E (2πε₀r²)/A
For an elemental point on the surface,
A = 2πrl = 2πr²
So,
η = -E ε₀
Hope this Helps!!!
what is meaning of convection
Answer:
Convection is heat transfer through the movement of liquids and gases.
1. A current of 0.001 A can be felt by the human body. 0.005 A can produce a pain response. 0.015 A can cause a loss of muscle control. In the procedures of this lesson, over 0.030 A of current traveled in the three-battery circuit. Why was this circuit safe to handle with dry hands?
Answer:
It was safe to handle the circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
Explanation:
Given;
Current of 0.001 A to be felt
Current of 0.005 A can produce a pain response
Current of 0.015 A can cause a loss of muscle control
Total current that traveled in the three-battery circuit = 0.03 A
Thus, we can conclude that, it was safe to handle the above mentioned circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves
O capture, amplify, and demodulate waves
change the amplitude and frequency of waves
O change the pulse and phase of waves
Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:
The diagram shows a device that uses radio waves.
What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves capture, amplify, and demodulate waves change the amplitude and frequency of waves change the pulse and phase of wavesAnswer:
2. capture, amplify, and demodulate waves
Explanation:
The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.
The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.
Hence, the correct option is 2.
Answer:
B
Explanation:
edge 2020
A noisy channel needs to transfer 87 kbps, but has a SNR of 11 dB (decibels). Calculate the minimum Bandwidth required , in kHz, according to Shannon.
Answer:
24KHz
Explanation:
See attached file
What is the electric potential (relative to infinity) due to these charges at the center of this square
Answer:
Zero
Explanation:
See attached file pls
Why do some astronomers object to the new definition of a planet that was adopted in 2006?
A. New space missions show that Pluto is much larger than originally thought.
B. By this definition, Earth, Jupiter, and other planets should not be considered planets.
C. There was never a vote on whether to adopt the new definition or not.
D. It means that we now technically have over 100 planets.
Answer:
A. New space missions show that Pluto is much larger than originally thought.
Explanation:
The new definition of a planet that was adopted in 2006, defined planet as an object that orbits the sun, with sufficient mass to be round, not a satellite of another object, and has removed debris and small objects from the area around its orbit.
This new definition of a planet that was adopted in 2006, classified Pluto as "dwarf planet", because Pluto meets planetary criteria except that it has not cleared debris from its orbital neighborhood.
However, new Horizons spacecraft flew by Pluto in 2015, revealed that Pluto is much larger than originally thought
Therefore, the correct option is "A"
A. New space missions show that Pluto is much larger than originally thought.
Answer: it means that we now technically have over 100 planets
Explanation:
it’s not New space missions show that Pluto is much larger than originally thought!!!
A grating having 5000 lines/cm is used with light of wavelength 633 nm. How many total maxima (count central maxima plus all those on either side of the central maxima) are produced
Answer:
The total number of maxima produced is [tex]m_T = 7[/tex] maxima
Explanation:
From the question we are told that
The number of lines per cm is [tex]n = 5000 \ lines/cm[/tex]
The wavelength of the light is [tex]\lambda = 633 nm = 633 *10^{-9} \ m[/tex]
Now the distance between the lines is mathematically evaluated as
[tex]d = \frac{1}{n}[/tex]
substituting values
[tex]d = \frac{1}{5000}[/tex]
[tex]d = \frac{1 *10^{-2}}{5000}[/tex] N/B - this statement convert it from cm to m
[tex]d = 2 *10^{ -6} \ m[/tex]
Generally the condition for diffraction i mathematically represented as
[tex]dsin(\theta ) = m \lambda[/tex]
at maximum [tex]\theta = 90 ^o[/tex]
[tex]d sin (90) = \lambda m[/tex]
here m is the number of maxima
Thus making m the subject we have
[tex]m = \frac{d sin (90)}{ \lambda }[/tex]
So [tex]m = \frac{2*10^{-6} sin (90)}{ 633 *10^{-9}}[/tex]
[tex]m = 3.2[/tex]
=> m =3
Now the total number of maxima would include the bright fringe(3) and dark fringe (3) plus the central maxima (1)
Thus
[tex]m_T = 3 + 3 +1[/tex]
[tex]m_T = 7[/tex] maxima
A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω.
The induced current in the loop at the given dimension is 10.25 A.
The given parameters:
number of turns of the solenoid, n = 1200 turnsdiameter of the solenoid, d = 2.0 cmlength of the solenoid, L = 7.5 cmresistance of the loop, R = 0.032 Ω.current in solenoid, I = 1.3 Atime, t = 30 msThe area are of the solenoid is calculated as;
[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times 0.02^2}{4} = 0.000314 \ m^2[/tex]
The emf induced in the solenoid is calculated as;
[tex]emf = N\frac{d\phi}{dt} \\\\emf = N (\frac{BA}{t} )\\\\emf = N(\frac{\mu_0 NI \times A} {L\times t} )\\\\emf = \frac{N^2 \mu_0 I A}{Lt} \\\\emf = \frac{(1200)^2 \times (4\pi \times 10^{-7}) \times (1.3) \times (0.000314)}{0.075 \times 30\times 10^{-3}} \\\\emf = 0.328 \ V[/tex]
The induced current in the loop is calculated as follows;
[tex]I = \frac{emf}{R} \\\\I = \frac{0.328}{0.032} \\\\I = 10.25 \ A[/tex]
Thus, the induced current in the loop is 10.25 A.
"Your question is not complete, it seems to be missing the following information;"
A solenoid passes through the center of a wire loop, as shown in (Figure 1). The solenoid has 1200 turns, a diameter of 2.0 cm, and is 7.5 cm long. The resistance of the loop is 0.032 Ω.If the current in the solenoid is increased by 1.3 A in 30 ms, what is the induced current in the loop?
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PLEASE! PLEASE! PLEASE! HELP! I need this by Monday! It's a project DUE on Monday! I'm giving 75 points for the Brainiest :-)
Answer:
Please see below for all the numbers to be entered in the table:
Explanation:
Coaster World: F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J; V = 40/10 = 4 m/s
Wally: F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s
Elijah: F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s
George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s
Answer:
Down below
Explanation:
Coaster World: F = 160 N; D = 40 m; T = 10 s; W = 160 * 40 = 6400 J; V = 40/10 = 4 m/s
Wally: F = 800 N; D = 10 m; T = 3.5 s W = 800*10 = 8000 J; V = 10/3.5 = 2.86 m/s
Elijah: F = 1400 N; D = 800 m; T = 40 m = 2400 s; W = 1400*800 = 112000 J; V = 800/2400 = 0.33 m/s
George: F = 600 N; D = 80 m; T = 50 m = 3000 s; W = 600 * 80 = 48000 j so he should get paid: 48,000/1000= $48; V = 80/3000 = 0.027 m/s
metal sphere A has 4 units of negative charge and metal sphere B has 2 units of positive charge. The two spheres are brought into contact. What is the final charge state of each sphere
In final state, every metal sphere carries 1 unit of negative charge.
When the spheres brought into contact, the charges try to get an equilibrium distribution on both spheres.It means that same amount of positive charges neutralise the same amount of negative charges.Given that, metal sphere A has 4 units of negative charge and metal sphere B has 2 units of positive charge. So that, 2 units of negative charge will neutralize the 2 unit of positive charge of the second sphere. The others 2 units of negative charge of the first sphere will distribute equally on both spheres.Hence, In final state, every metal sphere carries 1 unit of negative charge.
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A woman who weighs 500 N stands on an 8 m long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times bigger than the support force at the left end. How far from the right end is the woman standing
Answer:
[tex]1.6\; \rm m[/tex].
Explanation:
Let [tex]x[/tex] denote the distance (in meters) between the person and the right end of the board.
To keep the calculations simple, consider another unknown: let [tex]y[/tex] denote the support force (in Newtons) on the left end. The support force on the right end of this board would be [tex]3 \, y[/tex] (also in Newtons.)
Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)
In the first diagram, the support at the left end of the board is considered as the fulcrum. In the second diagram, the support at the right end of the board is considered as the fulcrum.Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to [tex]r[/tex], the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its [tex]r[/tex] will be zero, and it will have no torque in this context.
Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:
The weight of the board acts at the center of the board, [tex](1/2) \times 8\; \rm m = 4\; \rm m[/tex] from the fulcrum. This force will exert a torque of [tex]\tau(\text{weight of board}) = 4\; \rm m \times (-100\; \rm N) = (-400\; \rm N \cdot m)[/tex] on this "lever". The negative sign indicates that this torque points downwards.The weight of the person acts at [tex]x\;\rm m[/tex] from the right end of the board, which is [tex](8 - x)\; \rm m[/tex] from the fulcrum at the other end of this board. This force will exert a torque of [tex]\tau(\text{weight of person}) = (8 - x)\; {\rm m \times (-500\; \rm N)} = (-500\, \mathnormal{(8 - x)})\; \rm N \cdot m[/tex] on this "lever". This torque also points downwards.The support on the right end of the board acts at [tex]8\; \rm m[/tex] from the fulcrum (i.e., the left end of this board.) This force will exert a torque of [tex]\tau(\text{support, right}) = 8\; {\rm m} \times (3\, \mathnormal{y})\; {\rm N} = (24\, y)\; \rm N \cdot m[/tex] on the "lever". This torque points upwards.If the value of [tex]x[/tex] and [tex]y[/tex] are correct, these three torques should add up to zero. That is:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0[/tex].
That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:
[tex]\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0[/tex].
Combine these two equations into a two-by-two system. Solve the system for [tex]x[/tex] and [tex]y[/tex]:
[tex]\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.[/tex].
In other words, the person is standing at about [tex]1.6\; \rm m[/tex] from the right end of the board. The support force at the left end of the board is [tex]150\; \rm N[/tex].
After fixing a flat tire on a bicycle you give the wheel a spin. Its initial angular speed was 5.45 rad/s and it rotated 14.4 revolutions before coming to rest.
1. What was its average angular acceleration?
2. For what length of time did the wheel rotate?
Answer:
(a) α = -0.16 rad/s²
(b) t = 33.2 s
Explanation:
(a)
Applying 3rd equation of motion on the circular motion of the tire:
2αθ = ωf² - ωi²
where,
α = angular acceleration = ?
ωf = final angular velocity = 0 rad/s (tire finally stops)
ωi = initial angular velocity = 5.45 rad/s
θ = Angular Displacement = (14.4 rev)(2π rad/1 rev) = 28.8π rad
Therefore,
2(α)(28.8π rad) = (0 rad/s)² - (5.45 rad/s)²
α = -(29.7 rad²/s²)/(57.6π rad)
α = -0.16 rad/s²
Negative sign shows deceleration
(b)
Now, we apply 1st equation of motion:
ωf = ωi + αt
0 rad/s = 5.45 rad/s + (-0.16 rad/s²)t
t = (5.45 rad/s)/(0.16 rad/s²)
t = 33.2 s
Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mule and refuses to continue on the walk. Cass yanks on the chain with a 67 N force at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the tension force.
Answer:
Horizontal component: [tex]F_x = 58\ N[/tex]
Vertical component: [tex]F_y = 33.5\ N[/tex]
Explanation:
To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.
Therefore, for the horizontal component, we have:
[tex]F_x = F * cos(angle)[/tex]
[tex]F_x = 67 * cos(30)[/tex]
[tex]F_x = 58\ N[/tex]
For the vertical component, we have:
[tex]F_y = F * sin(angle)[/tex]
[tex]F_y = 67 * sin(30)[/tex]
[tex]F_y = 33.5\ N[/tex]
So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.
when electricity is transmitted through the air the resulting spark is called ?
Answer:
Lightning
Explanation:
Air is a non-conductor of electricity because it does not have free electrons to carry the current. But when there is a high voltage, as in the case of lightning, the molecules of air get ionized and electrons are available to flow to make the current and conduct electricity through the air.
Answer:
corona discharge
Explanation:
A corona discharge is an electrical discharge caused by the ionization of a fluid such as air, surrounding a conductor that is electrically charged.
A mass on a spring vibrates in simple harmonic motion at a frequency of 3.26 Hz and an amplitude of 5.76 cm. If the mass of the object of 0.218 kg, what is the spring constant
Answer:
91.48N/m
Explanation:
In a spring-mass system undergoing a simple harmonic motion, the inverse of the frequency f, of oscillation is proportional to the square root of the mass m, and inversely proportional to the square root of the spring constant, k. This can be expressed mathematically as follows;
[tex]\frac{1}{f}[/tex] = [tex]2\pi\sqrt{\frac{m}{k} }[/tex] -----------(i)
From the question;
f = 3.26 Hz
m = 0.218kg
Substitute these values into equation (i) as follows;
[tex]\frac{1}{3.26}[/tex] = [tex]2\pi\sqrt{\frac{0.218}{k} }[/tex] [Square both sides]
([tex]\frac{1}{3.26}[/tex])² = ([tex]2\pi[/tex])²([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = [tex]4\pi[/tex]²([tex]\frac{0.218}{k}[/tex]) [Take [tex]\pi[/tex] to be 3.142]
([tex]\frac{1}{10.6276}[/tex]) = [tex]4(3.142)[/tex]²([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = [tex]39.488[/tex]([tex]\frac{0.218}{k}[/tex])
([tex]\frac{1}{10.6276}[/tex]) = ([tex]\frac{8.608}{k}[/tex]) [Switch sides]
([tex]\frac{8.608}{k}[/tex]) = ([tex]\frac{1}{10.6276}[/tex]) [Re-arrange]
([tex]\frac{k}{8.608}[/tex]) = ([tex]\frac{10.6276}{1}[/tex]) [Cross-multiply]
k = 8.608 x 10.6276
k = 91.48N/m
Therefore, the spring constant of the spring is 91.48N/m
The bulldog and skateboard have a combined mass of 20 kg. In case B (the middle of the three pictures of the bulldog and the well), the bulldog and skateboard have a KE of 380 J at the bottom of the well. How deep is the well in meters?
Answer:
h = 1.94 m
Explanation:
When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:
Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board
K.E = P.E
K.E = mgh
h = K.E/mg
where,
h = depth of well = ?
K.E = Kinetic Energy at bottom = 380 J
m = mass of bull dog and skate board = 20 kg
g = 9.8 m/s²
Therefore,
h = 380 J/(20 kg)(9.8 m/s²)
h = 1.94 m
The linear charge density on the inner conductor is and the linear charge density on the outer conductor is
Complete Question
The complete question is shown on the first uploaded image (reference for Photobucket )
Answer:
The electric field is [tex]E = -1.3 *10^{-4} \ N/C[/tex]
Explanation:
From the question we are told that
The linear charge density on the inner conductor is [tex]\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m[/tex]
The linear charge density on the outer conductor is
[tex]\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m[/tex]
The position of interest is r = 37.3 mm =0.0373 m
Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )
Generally according to Gauss Law
[tex]E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}[/tex]
=> [tex]E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}[/tex]
substituting values
[tex]E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}[/tex]
[tex]E = -1.3 *10^{-4} \ N/C[/tex]
The negative sign tell us that the direction of the electric field is radially inwards
=> [tex]|E| = 1.3 *10^{-4} \ N/C[/tex]