Answer:
For red light= 39.7°
Blue light 39.2°
Explanation:
Given that refractive index for red light is 1.33 and that of blue light is 1.342
So angle of refraction for red light will be
Sinစi/ (sinစ2) =( nw)r/ ni
Sin 58° x 1.000293/1.33. =( sinစ2)r
0.64= sinစ2r
Theta2r = 39.7°
For blue light
Sinစi/ (sinစ2) =( nw)b/ ni
Sin 58° x 1.000293/1.342 =( sinစ2)b
0.632= sinစ2r
Theta2b= 39.19°
For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.
a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?
Answer:
Explanation:
(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)
ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .
ΔK = 0
ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .
ΔUg = positive
ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .
ΔUs = 0
ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence
ΔEth = negative .
b )
work done by athlete
= 400 x 2 = 800 J
energy output = 800 J
c )
It is 25% of metabolic energy output of his body
so metalic energy output of body
= 4x 800 J .
3200 J
power = energy output / time
= 3200 / 1.6
= 2000 W .
d )
1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .
2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .
A) Applying the energy equation
The positive terms is : ΔUg The negative terms is : ΔEth The zero term are : ΔK and ΔUsB) The energy output by the athlete is ; 800 Joules
C) The metabolic power is : 2000 w
D) When he performs the task in 1.2 s
The metabolic energy he expends is : the same His metabolic power is : moreGiven data :
Weight of barbell = 400 N
Height = 2.0 m
Time = 1.6 secs
efficiency of the human body = 25%
Speed = constant
A) From the energy equation the ΔK is zero because the athlete is lifting the barbell at a constant speed. ΔUg is positive because as the weight is lifted its potential energy increases. ΔEth ( change in energy of earth ) is negative because it exerts a force in opposite direction to displacement
B) Determine the energy output of the athlete
weight of barbell * Height = 400 * 2 = 800 J
C) Determine the metabolic power
Metabolic power = energy output / Time
where ; energy output = 4 * 800 = 3200
∴ Metabolic power = 3200 / 1.6
= 2000 w
D) When performs same task at 1.2 s
The metabolic energy he expends is the same and His metabolic power is more
Hence we can conclude that the answers to your questions are as listed above
Learn more : https://brainly.com/question/17807740
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]
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 26 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?
Answer:
The velocity is [tex]v_h = 19.2 \ m/s[/tex]
Explanation:
From the question we are told that
The speed of the roller coaster at ground level is [tex]v = 26 \ m/s[/tex]
Generally we can define the roller coaster speed at ground level using the an equation of motion as
[tex]v^2 = u^2 + 2 g s[/tex]
u is zero given that the roller coaster started from rest
So
[tex]26^2 = 0 + 2 * g * s[/tex]
So
[tex]s = \frac{26^2}{ 2 * g }[/tex]
=> [tex]s = 37.6 \ m[/tex]
Now the displacement half way is mathematically represented as
[tex]s_{h} = \frac{37.6}{2}[/tex]
[tex]s_{h} = 18.8 \ m[/tex]
So
[tex]v_h ^2 = u^2 + 2 * g * s_h[/tex]
Where [tex]v_h[/tex] is the velocity at the half way point
=> [tex]v_h = \sqrt{ 0 + 2 * 9.8 * 18.8 }[/tex]
=> [tex]v_h = 19.2 \ m/s[/tex]
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
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
A student is hammering a nail into a board. Where should he hold the hammer and why?
Answer:
At the end of the handle farthest from the head of the hammer.
Explanation:
The force of the hammer is greatest the longer the radius is on a which would be the length of the handle. Simple mechanical advantage.
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]
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]
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
Which of the following describes wavelength?
A.
the height of a wave
B.
the distance between crests of adjacent waves
C.
the distance a wave travels in a given amount of time
D.
the number of waves that pass a point in a given amount of time
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
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
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 384 Hz tuning fork produces standing waves with a wavelength of 0.90 m inside a resonance tube. The speed of sound at experimental conditions is
Answer:
v = 345.6m/s
Explanation:
v = 384 x 0.9 = 345.6
v = 345.6m/s
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 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
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 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:
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
hat a 15 kg body is pulled along a horizontal fictional table by a force of 4N what is the acceleration of the body
Answer:
Acceleration of the body is:
[tex]a=0.27\,\,m/s^2[/tex]
Explanation:
Use Newton's second Law to solve for the acceleration:
[tex]F=m\,\,a\\a=\frac{F}{m} \\a=\frac{4\,N}{15\,\,kg} \\a=0.27\,\,m/s^2[/tex]
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.
Two objects, one of mass m and the other of mass 2m, are dropped from the top of a building. When they hit the ground:_______.
a) the heavier one will have four times the kinetic energy of the lighter one.
b) the heavier one will have twice the kinetic energy of the lighter one.
c) the heavier one will have times the kinetic energy of the lighter one.
d) both of them will have the same kinetic energy.
Answer:
b) the heavier one will have twice the kinetic energy of the lighter one.
Explanation:
The kinetic energy of object with mass, m
K.E₁ = ¹/₂mv²
where;
m is mass of the object
v is the velocity of the object
Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate
The kinetic energy of object with mass, 2m
K.E₂ = ¹/₂(2m)v²
K.E₂ = 2(¹/₂mv²)
BUT K.E₁ = ¹/₂mv²
K.E₂ = 2(K.E₁)
Therefore, the heavier one will have twice the kinetic energy of the lighter one.
b) the heavier one will have twice the kinetic energy of the lighter one.
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
The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum value of the magnetic field in the wave
Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (in V) is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms? V †
Answer:
The average emf induced in the coil is 175 mV
Explanation:
Given;
number of turns of the coil, N = 1060 turns
diameter of the coil, d = 20.0 cm = 0.2 m
magnitude of the magnetic field, B = 5.25 x 10⁻⁵ T
duration of change in field, t = 10 ms = 10 x 10⁻³ s
The average emf induced in the coil is given by;
[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]
where;
A is the area of the coil
A = πr²
r is the radius of the coil = 0.2 /2 = 0.1 m
A = π(0.1)² = 0.03142 m²
[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]
Therefore, the average emf induced in the coil is 175 mV
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations was 0.15 m, its wavelength was two meters, and the period was 2/15 s. If a point on the wave at a specific time has a displacement of 0.12 m, what is the transverse speed of that point?
Answer:
15m/sExplanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;
[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]
Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = [tex]\frac{1}{(2/15)}[/tex]
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength [tex]\lambda[/tex] = 2m
Transverse speed [tex]v = f \lambda[/tex]
[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]
Hence, the transverse speed at that point is 15m/s
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]
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 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
Krishna and Seldon now try a homework problem. A policeman sitting in his unmarked police car sees an approaching motorcyclist go through a red light two blocks away. He turns on his siren at a frequency of 1000 Hz as the motorcyclist heads directly toward him at 61 mph (27.27 m/s). What frequency does the motorcyclist hear? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz What frequency does the motorcyclist hear when stopped with the police car approaching at 61 mph (27.27 m/s)? (Enter your answer to at least the nearest integer. Assume the speed of sound in air is 331 m/s.) Hz
Answer:
Explanation:
We shall apply formula of Doppler's effect
Here source is fixed and observer is approaching the source
f = f₀ x [(V + v ) / V ]
f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .
f = 1000 x [(331 + 27.27 ) / 331 ]
= 1082 .4 Hz
This is the frequency heard by motorcyclist .
When police car is approaching him when he is stopped
f = f₀ x [V /(V - v ) ]
v is velocity of police car .
= 1000 x 331 / (331 - 27.27)
= 1090 Hz
