Answer:
[tex]\theta=10.20^{\circ}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Explanation:
First of all, we analyze the system of blocks before starting to move.
[tex]\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0[/tex]
[tex]\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0[/tex]
[tex]16sin(\theta)-2.91cos(\theta)=0[/tex]
[tex]tan(\theta)=0.18[/tex]
[tex]\theta=arctan(0.18)[/tex]
[tex]\theta=10.20^{\circ}[/tex]
Hence, the incline angle θ for which both blocks begin to slide is 10.20°.
Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.
[tex]P_{A}sin(\theta)-F_{fA}-F_{spring}=0[/tex]
Where:
[tex]F_{spring} = k\Delta l=2.1\Delta l[/tex]
[tex]P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0[/tex]
[tex]\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}[/tex]
[tex]\Delta l=0.10 ft[/tex]
Therefore, the required stretch or compression in the connecting spring is 0.10 ft.
I hope it helps you!
(a) The inclined angle for which both blocks begin to slide is 10.3⁰.
(b) The compression of the spring is 0.22 ft.
The given parameters;
mass of block A, = 11 lbmass of block B, = 5 lbcoefficient of static friction for A, = 0.16coefficient of static friction for B, = 0.23 spring constant, k = 2.1 lb/ftThe normal force on block A and B:
[tex]F_n_A = m_Agcos \ \theta\\\\F_n_B = m_Bgcos \ \theta[/tex]
The frictional force on block A and B:
[tex]F_f_A = \mu_s_AF_n_A \\\\F_f_B = \mu_s_BF_n_A[/tex]
The net force on the blocks when they starts sliding;
[tex](m_Ag sin \theta+ m_Bgsin\theta) - (F_f_A + F_f_B) = 0\\\\m_Ag sin \theta+ m_Bgsin\theta = F_f_A + F_f_B\\\\m_Ag sin \theta+ m_Bgsin\theta = \mu_Am_Agcos\theta \ + \ \mu_Bm_Bgcos\theta\\\\gsin\theta(m_A + m_B) = gcos\theta (\mu_Am_A + \mu_Bm_B)\\\\\frac{sin\theta}{cos \theta} = \frac{\mu_Am_A\ + \ \mu_Bm_B}{m_A\ + \ m_B} \\\\tan\theta = \frac{(0.16\times 11) \ + \ (0.23 \times 5)}{11 + 5} \\\\tan\theta = 0.1819\\\\\theta = tan^{-1}(0.1819)\\\\\theta = 10.3 \ ^0[/tex]
The change in the energy of the blocks is the work done in compressing the spring;
[tex]\Delta E = W\\\\F_A (sin \theta )d- \mu F_n d= \frac{1}{2} kd^2\\\\F_A sin\theta \ - \ \mu F_A cos\theta = \frac{1}{2} kd\\\\d = \frac{2F_A(sin\theta - \mu cos \theta) }{k} \\\\d = \frac{2\times 11(sin \ 10.3\ - \ 0.16\times cos \ 10.3) }{2.1} \\\\d = 0.22 \ ft[/tex]
Learn more here:https://brainly.com/question/16892315
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field to have a maximum value of the induced emf equal to 8.0 V
Answer:
The frequency of the coil is 7.07 Hz
Explanation:
Given;
number of turns of the coil, 200 turn
cross sectional area of the coil, A = 300 cm² = 0.03 m²
magnitude of the magnetic field, B = 30 mT = 0.03 T
Maximum value of the induced emf, E = 8 V
The maximum induced emf in the coil is given by;
E = NBAω
Where;
ω is angular frequency = 2πf
E = NBA(2πf)
f = E / 2πNBA
f = (8) / (2π x 200 x 0.03 x 0.03)
f = 7.07 Hz
Therefore, the frequency of the coil is 7.07 Hz
A 300 MWe (electrical power output) Power Plant having a thermal efficiency of 40% is cooled by sea water. Due to environmental regulations the seawater can only increase temperature by 5 C during the process. How much sea water (minimum) must be used in kg/s for cooling if the plant operates at it's rated capacity?
Answer:
m = 22,877 kg / s
Explanation:
Let's solve this exercise in parts, first look for the amount of heat generated by the plant and then the amount of water to dissipate this heat
The plant generates a power of 300 MW at a rate of 40%, let's use a direct ratio rule to find the heat. If the power is 400 MW it corresponds to 40%, what heat (Q) corresponds to the other 60%
Q = 300 60% / 40%
Q = 450 MW
having the amount of heat generated we can use the calorimeter equation,
Q = m [tex]c_{e}[/tex] [tex](T_{f} - T_{o})[/tex]
m = Q / c_{e} (T_{f} - T_{o})
let's use the maximum temperature change allowed
(T_{f} - T_{o}) = 5
the specific heat of sea water is 3934 J / kg ºC, note that it is less than that of pure water, due to the salts dissolved in sea water
power and energy are related
W = Q / t
Q = W t
let's calculate
m = 450 10⁶ / (3934 5)
m = 22,877 kg / s
The concentration of sodium and potassium ions in the blood and body fluid is regulated by :
Answer:
Kidney
Explanation:
One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.
Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.
Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.
A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.
Explanation:
EE = ½ kx²
EE = ½ (800 N/m) (6 m)²
EE = 14,400 J
Answer:
14,400 J
Explanation:
Its the answer
A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is so large that the ball flies up at an angle of 30 degrees above the ground. The player however claims (s)he aimed her/his foot at a 40 degree angle above the ground. Calculate the average kicking force magnitude and the final speed of the ball, if you are given that the foot was in contact with the ball for one hundredth of a second.
Answer:
v_{f} = 74 m/s, F = 230 N
Explanation:
We can work on this exercise using the relationship between momentum and moment
I = ∫ F dt = Δp
bold indicates vectors
we can write this equations in its components
X axis
Fₓ t = m ( -v_{xo})
Y axis
t = m (v_{yf} - v_{yo})
in this case with the ball it travels horizontally v_{yo} = 0
Let's use trigonometry to write the final velocities and the force
sin 30 = v_{yf} / vf
cos 30 = v_{xf} / vf
v_{yf} = vf sin 30
v_{xf} = vf cos 30
sin40 = F_{y} / F
F_{y} = F sin 40
cos 40 = Fₓ / F
Fₓ = F cos 40
let's substitute
F cos 40 t = m ( cos 30 - vₓ₀)
F sin 40 t = m (v_{f} sin 30-0)
we have two equations and two unknowns, so the system can be solved
F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)
F sin 40 0.1 = 0.4 v_{f} sin 30
we clear fen the second equation and subtitles in the first
F = 4 sin30 /sin40 v_{f}
F = 3.111 v_{f}
(3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80
v_{f} (3,111 cos 40 -4 cos30) = - 80
v_{f} (- 1.0812) = - 80
v_{f} = 73.99
v_{f} = 74 m/s
now we can calculate the force
F = 3.111 73.99
F = 230 N
A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)
Answer:I don’t know
Explanation:
The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps and dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 1/2 in).
If we treat the spring assembly as a single spring, what is the approximate spring constant?
k= ____________
Answer:
The approximate spring constant is [tex]k = 55533.33 \ N/m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 68 \ kg[/tex]
The dip of the car is [tex]x = 1.2 \ cm = 0.012 \ m[/tex]
Generally according to hooks law
[tex]F = k * x[/tex]
here the force F is the weight of the person which is mathematically represented as
[tex]F = m * g[/tex]
=> [tex]m * g = k * x[/tex]
=> [tex]k = \frac{m * g }{x }[/tex]
=> [tex]k = \frac{68 * 9.8}{ 0.012}[/tex]
=> [tex]k = 55533.33 \ N/m[/tex]
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
A sinusoidal voltage Δv = (100 V) sin (170t) is applied to a series RLC circuit with L = 40 mH, C = 130 μF, and R = 50 Ω.
Required:
a. What is the impedance of the circuit?
b. What is the maximum current in the circuit?
Answer:
See attached file
Explanation:
What is the emf of this cell under standard conditions? Express your answer using three significant figures.
Complete Question
A voltaic cell utilizes the following reaction and operates at 298 K:
3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).
What is the emf of this cell under standard conditions? Express your answer using three significant figures.
Answer:
The value is [tex]E^o_{cell} = 2.35 V[/tex]
Explanation:
From the question we are told that
The ionic equation is
[tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]
Now under standard conditions the reduction half reaction is
[tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]
And the oxidation half reaction is
[tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]
The emf of this cell under standard conditions is mathematically represented as
[tex]E^o_{cell} = E^o _r - E^o _o[/tex]
substituting values
[tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]
[tex]E^o_{cell} = 2.35 V[/tex]
Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the width of the central maximum on a screen 1.0 m from the slit?
Answer:
The width is [tex]w_c = 0.00422 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 6.33*10^{-7} \ m[/tex]
The width of the slit is [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 1.0 \ m[/tex]
Generally the central maximum is mathematically represented as
[tex]w_c = 2 * y[/tex]
Here y is the width of the first order maxima which is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]
[tex]y = 0.00211 \ m[/tex]
So
[tex]w_c = 2 *0.00211[/tex]
[tex]w_c = 0.00422 \ m[/tex]
At sea level, at a latitude where , a pendulum that takes 2.00 s for a complete swing back and forth has a length of 0.993 m. What is the value of g in m/s2 at a location where the length of such a pendulum is 0.970 m
Answer:
a) The value of g at such location is:
[tex]g=9.8005171\,\frac{m}{s^2}[/tex]
b) the period of the pendulum with the length is 0.970 m is:
[tex]T=1.9767 sec[/tex]
Explanation:
Recall the relationship between the period (T) of a pendulum and its length (L) when it swings under an acceleration of gravity g:
[tex]L=\frac{g}{4\,\pi^2} \,T^2[/tex]
a) Then, given that we know the period (2.0 seconds), and the pendulum's length (L=0.993 m), we can determine g at that location:
[tex]g=\frac{4\,\pi^2\,L}{T^2}\\g=\frac{4\,\pi^2\,0.993}{(2)^2}\\g=\pi^2\,(0.993)\,\frac{m}{s^2} \\g=9.8005171\,\frac{m}{s^2}[/tex]
b) for this value of g, when the pendulum is shortened to 0.970 m, the period becomes:
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by . where at one end of the rope, is in meters, and is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?
Complete question is;
A rope, under a tension of 153 N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by
y = (0.15 m) sin[πx/3] sin[12π t].
where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c)the mass of the rope? (d) If the rope oscillates in a third - harmonic standing wave pattern, what will be the period of oscillation?
Answer:
A) Length of rope = 4 m
B) v = 24 m/s
C) m = 1.0625 kg
D) T = 0.11 s
Explanation:
We are given;
T = 153 N
y = (0.15 m) sin[πx/3] sin[12πt]
Comparing this displacement equation with general waveform equation, we have;
k = 2π/λ = π/2 rad/m
ω = 2πf = 12π rad/s
Since, 2π/λ = π/2
Thus,wavelength; λ = 4 m
Since, 2πf = 12π
Frequency;f = 6 Hz
A) We are told the rope oscillates in a second-harmonic standing wave pattern. So, we will use the equation;
λ = 2L/n
Since second harmonic, n = 2 and λ = L = 4 m
Length of rope = 4 m
B) speed is given by the equation;
v = fλ = 6 × 4
v = 24 m/s
C) To calculate the mass, we will use;
v = √T/μ
Where μ = mass(m)/4
Thus;
v = √(T/(m/4))
Making m the subject;
m = 4T/v²
m = (4 × 153)/24²
m = 1.0625 kg
D) Now, the rope oscillates in a third harmonic.
So n = 3.
Using the formula f = 1/T = nv/2L
T = 2L/nv
T = (2 × 4)/(3 × 24)
T = 0.11 s
For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?
Answer: 56°
Explanation:
Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.
The formula is;
[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]
[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]
= 56.30993247
= 56°
A 1.5 V battery is connected to a 1000 ohm resistor and a 500 ohm resistor in series. The voltage across the 1000 ohm resistor is _____ V.
Answer:
1 volt and 0.5 voltExplanation:
Given data
voltage supplied Vs= 1.5 volts
resistance R1= 1000 ohms
resistance R2= 500 ohms
The total resistance is
Rt= 1000+ 500
Rt= 1500 ohms
The current I is given as
[tex]I= \frac{Vs}{Rt} \\\\ I= \frac{1.5}{1500} = 0.001mA[/tex]
Voltage across R1
[tex]VR1= Vs(\frac{R1}{R1+R2} )=1.5(\frac{1000}{1000+500} )= 1.5(\frac{1000}{1500} )\\ \\\ VR1= 1v[/tex]
Voltage across R2
[tex]VR2= Vs(\frac{R2}{R1+R2} )=1.5(\frac{500}{1000+500} )= 1.5(\frac{500}{1500} ) \\\ VR2=0.5v[/tex]
In series connection the current is the same for all components while the voltage divides across all components,the voltages consumed by each individual resistance is equal to the source voltage.
In the summer of 2010 a huge piece of ice roughly four times the area of Manhattan and 500 m thick caved off the Greenland mainland.
Required:
a. How much heat would be required to melt this iceberg (assumed to be at 0°C) into liquid water at 0°C?
b. The annual U.S. energy consumption is 1.2 x 10^20 J. If all the U.S. energy was used to melt the ice, how many days would it take to do so?
Answer:
a
[tex]Q = 5.34 *10^{19} \ J[/tex]
b
[tex]T = 0.445 * 365 = 162. 413 \ days[/tex]
Explanation:
From the question we are told that
The area of Manhattan is [tex]a_k = 87.46 *10^{6} \ m^2[/tex]
The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]
The thickness is [tex]t = 500 \ m \\[/tex]
Generally the volume of the ice is mathematically represented is
[tex]V = a_i * t[/tex]
substituting value
[tex]V = 500 * 3.498*10^{8}[/tex]
[tex]V = 1.75 *10^{11}\ m^3[/tex]
Generally the mass of the ice is
[tex]m_i = \rho_i * V[/tex]
Here [tex]\rho_i[/tex] is the density of ice the value is [tex]\rho _i = 916.7 \ kg/m^3[/tex]
=> [tex]m_i = 916.7 * 1.75*10^{11}[/tex]
=> [tex]m_i = 1.60 *10^{14} \ kg[/tex]
Generally the energy needed for the ice to melt is mathematically represented as
[tex]Q = m _i * H_f[/tex]
Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is [tex]H_f = 3.33*10^{5} \ J/kg[/tex]
=> [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]
=> [tex]Q = 5.34 *10^{19} \ J[/tex]
Considering part b
We are told that the annual energy consumption is [tex]G = 1.2*10^{20 } \ J / year[/tex]
So the time taken to melt the ice is
[tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]
[tex]T = 0.445 \ years[/tex]
converting to days
[tex]T = 0.445 * 365 = 162. 413 \ days[/tex]
g One of the harmonics in an open-closed tube has frequency of 500 Hz. The next harmonic has a frequency of 700 Hz. Assume that the speed of sound in this problem is 340 m/s. a. What is the length of the tube
Answer:
The length of the tube is 85 cm
Explanation:
Given;
speed of sound, v = 340 m/s
first harmonic of open-closed tube is given by;
N----->A , L= λ/₄
λ₁ = 4L
v = Fλ
F = v / λ
F₁ = v/4L
Second harmonic of open-closed tube is given by;
L = N-----N + N-----A, L = (³/₄)λ
[tex]\lambda = \frac{4L}{3}\\\\ F= \frac{v}{\lambda}\\\\F_2 = \frac{3v}{4L}[/tex]
Third harmonic of open-closed tube is given by;
L = N------N + N-----N + N-----A, L = (⁵/₄)λ
[tex]\lambda = \frac{4L}{5}\\\\ F= \frac{v}{\lambda}\\\\F_3 = \frac{5v}{4L}[/tex]
The difference between second harmonic and first harmonic;
[tex]F_2 -F_1 = \frac{3v}{4L} - \frac{v}{4L}\\\\F_2 -F_1 = \frac{2v}{4L} \\\\F_2 -F_1 =\frac{v}{2L}[/tex]
The difference between third harmonic and second harmonic;
[tex]F_3 -F_2 = \frac{5v}{4L} - \frac{3v}{4L}\\\\F_3 -F_2 = \frac{2v}{4L} \\\\F_3 -F_2 =\frac{v}{2L}[/tex]
Thus, the difference between successive harmonic of open-closed tube is
v / 2L.
[tex]700H_z- 500H_z= \frac{v}{2L} \\\\200 = \frac{v}{2L}\\\\L = \frac{v}{2*200} \\\\L = \frac{340}{2*200}\\\\L = 0.85 \ m\\\\L = 85 \ cm[/tex]
Therefore, the length of the tube is 85 cm
Rank these electromagnetic waves on the basis of their speed (in vacuum). Rank from fastest to slowest.
a. Yellow light
b. FM radio wave
c. Green light
d. X-ray
e. AM radio wave
f. Infrared wave
Answer:
From fastest speed to slowest speed, the electromagnetic waves are ranked as(up to down):
d. X-ray
c. Green light
a. Yellow light
f. Infrared wave
b. FM radio wave
e. AM radio wave
Explanation:
Electromagnetic waves are waves produced as a result of vibrations between an electric field and a magnetic field. The waves have three properties and these properties are frequency, speed and wavelength, which are related by the relationship below
V = Fλ
where:\
V = speed (velocity)
F = frequency
λ = wavelength.
From the relationship above, it is seen that the speed of a wave is directly proportional to its frequency. The higher the frequency, the higher the speed. Therefore, from the list given, the waves with the highest to lowest frequencies/ from left to right are:
X-ray (3×10¹⁹ Hz to 3×10¹⁶Hz), Green light (5.66×10¹⁴Hz), Yellow light (5.17×10¹⁴Hz), Infrared wave (3×10¹¹Hz), FM radio wave (10.8×10⁸Hz to 8.8×10⁷Hz), AM radio wave (1.72 × 10⁶Hz to 5.5×10⁵Hz).
This corresponds to the speed from highest to lowest from left to right.
Monochromatic light of wavelength, λ is traveling in air. The light then strikes a thin film having an index of refraction n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minimum film thickness will result in minimum reflection of this light?
Answer:
tmin= lambda/2
Explanation:
See attached file pls
A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What happens after he pulls his arms inwards
Answer:
His angular velocity will increase.
Explanation:
According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.
The angular momentum of a system = [tex]I[/tex]'ω'
where
[tex]I[/tex]' is the initial rotational inertia
ω' is the initial angular velocity
the rotational inertia = [tex]mr'^{2}[/tex]
where m is the mass of the system
and r' is the initial radius of rotation
Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.
we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].
From
[tex]I[/tex]'ω' = [tex]I[/tex]ω
since [tex]I[/tex] is now reduced, ω will be greater than ω'
therefore, the angular velocity increases.
A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded as shown. Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct
Explanation:
the missing figure in the Question has been put in the attachment.
Then from the figure we can observe that
the center of the sphere is positive, therefore, negative charge will be induced at A.
As B is grounded there will not be any charge on B
Hence the answer is A is negative and B is charge less.
Two narrow slits are illuminated by a laser with a wavelength of 593 nm. The interference pattern on a screen located x = 4.80 m away shows that the fourth-order bright fringe is located y = 8.20 cm away from the central bright fringe. Calculate the distance between the two slits.
Answer:
The distance is [tex]d = 1.39 *10^{-4} \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 593 \ nm = 593 *10^{-9} \ m[/tex]
The distance of the screen is x = 4.80 m
The location of the fourth order bright fringe is y = 8.20 cm = 0.082 m
The order of the fringe is n = 4
Generally the position of a fringe with respect to the central fringe is mathematically represented as
[tex]y = \frac{ n * x * \lambda }{d}[/tex]
Where d is the distance between the slits, so making d the subject
[tex]d = \frac{\lambda * x * n }{ y }[/tex]
substituting values
[tex]d = \frac{ 593 *10^{-9} * 4.80 * 4 }{ 0.082 }[/tex]
[tex]d = 1.39 *10^{-4} \ m[/tex]
A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the peak current in this circuit is 0.8484 A, what is the impedance of this circuit?
A) 200 Ω
B) 141 Ω
C) 20.4 Ω
D) 120 Ω
E) 100 Ω
Answer:A 200
Explanation:
Vp=1.41*Vrms
Vp=169.7 v
Z=Vp/Ip
Z=169.7/.8484
Z=200.03 ohm
The location of a particle is measured with an uncertainty of 0.15 nm. One tries to simultaneously measure the velocity of this particle. What is the minimum uncertainty in the velocity measurement. The mass of the particle is 1.770×10-27 kg
Answer:
198 ms-1
Explanation:
According to the Heisenberg uncertainty principle; it is not possible to simultaneously measure the momentum and position of a particle with precision.
The uncertainty associated with each measurement is given by;
∆x∆p≥h/4π
Where;
∆x = uncertainty in the measurement of position
∆p = uncertainty in the measurement of momentum
h= Plank's constant
But ∆p= mΔv
And;
m= 1.770×10^-27 kg
∆x = 0.15 nm
Making ∆v the subject of the formula;
∆v≥h/m∆x4π
∆v≥ 6.6 ×10^-34/1.770×10^-27 × 1.5×10^-10 ×4×3.142
∆v≥198 ms-1
Why was Bohr's atomic model replaced by the
modern atomic model?
Answer:
Explanation:
Bohr's atomic model was replaced by the modern atomic model because of its limitations, which included :
(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )
(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).
(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.
There is a river in front of you that flows due South at 3.0m/s. You launch a toy boat across the river with the front of the boat pointed due East. When you tested the boat on a still pond, the boat moved at 4.0m/s. Now as it moves to the opposite bank, it travels at some speed relative to you, sitting in your chair. What is this speed
Answer:
5.0 m/sExplanation:
If the river moves towards the south at 3m/s and the both moves towards the east at 4.0m/s, the speed of the boat relative to me will be the resulting displacement of both velocities of the river and that of the boat. This can be gotten using pythagoras theorem.
Let Vr be the relative speed. According to the theorem;
[tex]V_r^2 = V_s^2 + V_e^2\\\\V_r^2 = 3.0^2 + 4.0^2\\\\V_r^2 = 9+16\\\\V_r^2 = \sqrt{25}\\ \\V_r = 5.0m/s[/tex]
Hence this relative speed is 5.0 m/s
a ball is kicked on level ground with a speed of 30 m/s at angle of 40 degrees above horizontal g Find the minimum velocity of the ball during its flight
Answer:
The minimum velocity of the ball during its flight is 22.98 m/s.
Explanation:
The velocity of the ball v = 30 m/s
The angle it makes with the horizontal ∅ = 40°
The minimum velocity of the ball during flight will be the horizontal axis component of the velocity, as acceleration is zero on this axis.
[tex]V_{x}[/tex] = v cos ∅
[tex]V_{x}[/tex] = 30 cos 40°
[tex]V_{x}[/tex] = 30 x 0.766 = 22.98 m/s
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.
Answer:
B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.
C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.
E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.
#FreeMelvin
The same force is applied to two hoops. The hoops have the same mass, but the larger hoop has twice the radius. How are the angular accelerations of the hoops related
Answer:
The angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex].
Explanation:
Net force on the hoop is given by;
[tex]F_{net} = ma[/tex]
where;
a is linear acceleration
m is the mass
Net torque on the hoop is given by;
[tex]\tau_{net} =I\alpha[/tex]
where;
I is moment of inertia
α is the angular acceleration
But, τ = Fr
[tex]Fr = I \alpha\\\\\alpha = \frac{Fr}{I} \\\\\alpha = \frac{Fr}{mr^2} \\\\\alpha = \frac{F}{mr} \\\\\alpha = \frac{1}{r} (\frac{F}{m} )\\\\(since\ the \ force\ and \ mass \ are \ the \ same, \frac{F}{m} = constant=k)\\\\ \alpha = \frac{k}{r}\\\\k = \alpha r[/tex]
[tex]\alpha _1 r_1= \alpha_2 r_2[/tex]
let the angular acceleration of the smaller hoop = α₁
let the radius of the smaller hoop = r₁
then, the radius of the larger loop, r₂ = 2r₁
let the angular acceleration of the larger hoop = α₂
[tex]\alpha _1 r_1= \alpha_2 r_2\\\\\alpha_2= \frac{ \alpha _1 r_1}{r_2} \\\\\alpha_2=\frac{\alpha _1 r_1}{2r_1} \\\\\alpha_2= \frac{\alpha _1}{2} \\\\\alpha _1 = 2\alpha_2[/tex]
Therefore, the angular accelerations of the hoops are related by the following equation [tex]\alpha _1 = 2\alpha_2[/tex]
