Answer:
Explanation:
wavelength of light λ = 502 x 10⁻⁹ m /s
screen distance D = 1.4 m
Slit separation d = ?
position of n the separation is given by the formula
x = n Dλ / d , n is order of fringe , x = distance of n th fringe
10.4 x 10⁻³ = 20 x 1.4 x 502 x 10⁻⁹ / d
d = 20 x 1.4 x 502 x 10⁻⁹ / 10.4 x 10⁻³
= 1351.54 x 10⁻⁶
= 1.35 x 10⁻³ m
1.35 mm.
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A. The angular velocity of A is twice that of B. The angular velocity of A is half that of B. The angular velocity of A equals that of B.
Answer:
The angular velocity of A equals that of B.
Explanation:
Since both coins are on the same turn table that is rotating with a given angular velocity about a particular axis, then the angular velocity of coin A will be equal to the angular velocity of coin B about that same axis.
A pilot flies in a straight path for 1 hour 30 minutes. She then makes a course correction, heading 45 degrees to the right of her original course, and flies 2 hours 15 minutes in the new direction. If she maintains a constant speed of 345 mi/h, how far is she from her starting position? Give your answer to the nearest mile.
Answer:
1199 miles
Explanation:
1 hour 30 minutes = 1 + 30/60 = 1.5 hours
2 hours 15 minutes = 2 + 15/60 = 2.25 hours
The distance she flew in the 1st segment is:
1.5*345 = 517.5 miles
The distance she flew in the 2nd segment is:
2.25 * 345 = 776.25 miles
Since the 2nd segment is 45 degree with respect to the 1st segment, this means that she has flown
776.25 * cos(45) = 549 miles in-line with the 1st segment and
776.25* sin(45) = 549 miles perpendicular to the 1st segment:
So the distance from the end to her starting position is
[tex]\sqrt{(517.5 + 549)^2 + 549^2} = 1199 miles[/tex]
How does the body maintain homeostasis when its inner temperature becomes higher than the normal body
temperature?
by shivering
by sweating
by producing goose bumps
Answer:
by sweating
Explanation:
An electron follows a helical path in a uniform magnetic field given by:B =(20i^−50j^−30k^)mTAt time t = 0, the electron's velocity is given by:⃗v=(40i^−30j^+50k^)m/sa. What is the angle ϕ between v and B. The electron's velocity changes with time. Do b. its speed c. the angled. What is the radius of the helical path?
Answer:
a) 1.38°
b) 7.53*10^11 m/s/s
c) 6.52*10^-9m
Explanation:
a) to find the angle you can use the dot product between two vectors:
[tex]\vec{v}\cdot\vec{B}=vBcos\theta\\\\\theta=cos^{1}(\frac{\vec{v}\cdot\vec{B}}{vB})[/tex]
v: velocity of the electron
B: magnetic field
By calculating the norm of the vectors and the dot product and by replacing you obtain:
[tex]B=\sqrt{(20)^2+(50)^2+(30)^2}=61.64mT\\\\v=\sqrt{(40)^2+(30)^2+(50)^2}=70.71m/s\\\\\vec{v}\cdot\vec{B}=[(20)(40)+(50)(30)-(30)(50)]mTm/s=800mTm/s\\\\\theta=cos^{-1}(\frac{800*10^{-3}Tm/s}{(70.71m/s)(61.64*10^{-3}T)})=cos^{-1}(0.183)=1.38\°[/tex]
the angle between v and B vectors is 1.38°
b) the change in the speed of the electron can be calculated by the change in the momentum in the following way:
[tex]\frac{dp}{dt}=F_e=qvBsin\theta\\\\\frac{dp}{dt}=(1.6*10^{-19}C)(70.71m/s)(61.64*10^{-3}T)(sin(1.38\°))=6.85*10^{-19}N[/tex]
due to the mass of the electron is a constant you have:
[tex]\frac{dp}{dt}=\frac{mdv}{dt}=6.85*10^{-19}N\\\\\frac{dv}{dt}=\frac{6.85*10^{-19}N}{9.1*10^{-31}kg}=7.53*10^{11}(m/s)/s[/tex]
the change in the speed is 7.53*10^{11}m/s/s
c) the radius of the helical path is given by:
[tex]r=\frac{m_ev}{qB}=\frac{(9.1*10^{-31}kg)(70.71m/s)}{(1.6*10^{-19}C)(61.64*10^{-3}T)}=6.52*10^{-9}m[/tex]
the radius is 6.52*10^{-9}m
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank in increasing order by placing the most negative on the left and the most positive on the right. To rank items as equivalent, overlap them.
The electric force between charges can be determined using Coulomb's law. Positive-positive and negative-negative combinations have repulsive forces, while positive-negative combinations have attractive forces. Neutral charges do not exert any force.
The electric force between two charges can be determined using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Using this information, we can rank the six combinations of electric charges based on the electric force acting on q1:
Combination 1: Positive charge (+) and positive charge (+) - repulsive forceCombination 2: Negative charge (-) and negative charge (-) - repulsive forceCombination 3: Positive charge (+) and negative charge (-) - attractive forceCombination 4: Negative charge (-) and positive charge (+) - attractive forceCombination 5: Positive charge (+) and neutral charge (0) - no forceCombination 6: Negative charge (-) and neutral charge (0) - no forceLearn more about Electric force here:
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The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horizontally from north to south. At the electron's latitude the vertical component of the Earth's magnetic field points down with a magnitude of 42.3 μT. What is the direction of the force on the electrons due to this component of the magnetic field?
Answer:
The direction is due south
Explanation:
From the question we are told that
The energy of the electron is [tex]E = 19.0keV = 19.0 *10^3 eV[/tex]
The earths magnetic field is [tex]B = 42.3 \muT = 42.3 *10^{-6} T[/tex]
Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as
[tex]\= F = q (\= v * \=B)[/tex]
On the first uploaded image is an illustration of the movement of the electron
Looking at the diagram we can see that in terms of direction the magnetic force is
[tex]\= F =q(\=v * \= B)= -( -\r i * - \r k)[/tex]
[tex]= -(- (\r i * \r k))[/tex]
generally i cross k = -j
so the equation above becomes
[tex]\= F = -(-(- \r j))[/tex]
[tex]= - \r j[/tex]
This show that the direction is towards the south
8. In the procedure for measuring the frequency of oscillation, we are instructed to pull downward on the hanging mass about 10 cm. If we performed the experiment a second time and deliberately pulled down on the hanging mass by 15 cm, would our period change? Justify your answer. (5 points)
Answer:
No
Explanation:
The frequency of oscillation of spring mass system is independent of its amplitude.
The frequency of oscillation will not change because it is not a function of amplitude.
What is the spring-mass system?In a real spring-mass system, the spring has a non-negligible mass m. Since not all of the spring's length moves at the same velocity v as the suspended mass M,
To determine the frequency of oscillation, the effective mass of the spring is defined as the mass that needs to be added to M to correctly predict the behaviour of the system.
For vertical springs, however, we need to remember that gravity stretches or compresses the spring beyond its natural length to the equilibrium position. After we find the displaced position
Thus the frequency of oscillation will not change because it is not a function of amplitude.
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A 113 ‑turn circular coil of radius 2.71 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 10.3 Ω resistor to create a closed circuit. During a time interval of 0.153 s, the magnetic field strength decreases uniformly from 0.441 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.
Answer:
357 mJ
Explanation:
Parameters given:
Number of turns, N = 113
Radius, r = 2.71 cm = 0.0271 m
Resistance connected to coil, R = 10.3 ohms
Time interval, dt = 0.153 s
Change in magnetic field, dB = 0 - 0.441 T = - 0.441 T
First, we need to find the induced EMF in the coil. It is given as:
V = - (N * A * dB)/dt
A = area of coil = pi * r² = 3.142 * 0.0271² = 0.0023 m²
Therefore:
V = - (113 * 0.0023 * - 0.441) / 0.153
V = 0.75 V
Electrical energy dissipated in the resistor is given as:
E = V² / (R * t)
Where t is the time (0.153 secs)
E = (0.75²) / (10.3 * 0.153)
E = 0.357 J = 357 mJ
The energy dissipated in the resistor is 357 mJ
Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds the other end loosely in his hand, so that the speed of waves on the clothesline is a relatively slow 0.740m/s . He finds several frequencies at which he can oscillate his end of the clothesline so that a light clothespin 43.0cm from the post doesn't move. What are these frequencies?
Answer:
Frequencies are 0.86Hz
Explanation:
Given v=0.74m/s
d= 0.43m
Lambda/2=0.43m
Thus lambda= 0.86m
And F= v/lambda
0.74/0.86= 0.86Hz
Answer:
The obsrved frquencies are [tex]f= n (0.86) Hz[/tex] n = 1,2,3,...,n.
Explanation:
From the question we are told that
The speed of the wave is v = 0.740 m/s
The distance from the post is [tex]d = 43.0cm = \frac{43}{10} = 0.43m[/tex]
Generally frequency is mathematically represented as
[tex]f = \frac{v}{\lambda }[/tex]
so for the first frequency which the the fundamental frequency(first harmonic frequency) the clothesline(string) would form only one loop and hence the length between the vertical post and throcky hands is mathematically represented as
[tex]L = \frac{\lambda}{2}[/tex]
=> [tex]\lambda = 2L[/tex]
So
[tex]f_f =f_1= \frac{v}{2L}[/tex]
Substituting values
[tex]f_f = f_1 = \frac{0.740}{2* 0.43}[/tex]
[tex]= 0.860 Hz[/tex]
For the second harmonic frequency i.e is when the clothesline(string) forms two loops the length is mathematically represented as
[tex]L = \lambda[/tex]
So the second harmonic frequency is
[tex]f_2 = \frac{v}{L}[/tex]
[tex]f_2 = 2 * \frac{v}{2L}[/tex]
[tex]f_2 =\frac{0.740}{0.430}[/tex]
[tex]= 1.72Hz[/tex]
For the third harmonic frequency i.e when the clothesline(string) forms three loops the length is mathematically represented as
[tex]L = 3 * \frac{\lambda }{2}[/tex]
So the wavelength is
[tex]\lambda = \frac{2L}{3}[/tex]
And the third harmonic frequency is mathematically evaluated as
[tex]f_3 = \frac{v}{\frac{2L}{3} }[/tex]
[tex]= 3 * \frac{v}{2L}[/tex]
[tex]f_3=2.58Hz[/tex]
Looking at the above calculation we can conclude that the harmonic frequency of a vibrating string(clothesline) can be mathematically represented as an integer multiple of the fundamental frequency
i.e
[tex]f = n f_f[/tex]
[tex]f = n \frac{v}{2L}[/tex]
[tex]f= n (0.86) Hz[/tex]
Where n denotes integer values i.e n = 1,2,3,....,n.
Note : The length that exist between two nodes which are successive is equivalent to half of the wavelength so when one loop is form the number of nodes would be 2 and on anti-node
Magnetic fields and electric fields are identical in that they both-
Answer:
Similarities between magnetic fields and electric fields: Electric fields are produced by two kinds of charges, positive and negative. Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like poles repel; unlike poles attract.
they attract each other
Explanation:
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it. Imagine that you slowly pull down with a spring scale so the block is now below the equilibrium position where it was hanging at rest. The scale reading when you let go of the block is 3.0 N.
Part A
Where was the block when you let go? Assume y0 is the equilibrium position of the block and that "down" is the positive direction.
Part B
Determine the work you did stretching the spring.
Express your answer to two significant figures and include the appropriate units.
Part C
What was the energy of the spring-Earth system when you let go (assume that zero potential energy corresponds to the equilibrium position of the block)?
Express your answer to two significant figures and include the appropriate units.
Part D
How far will the block rise after you release it?
Express your answer to two significant figures and include the appropriate units.
Answer:
a) x = 0.144 m
b) W = 0.15 J
c) E = 0.14 J
d) The block will rise 0.07m after it is released
Explanation:
a) The elastic force equals the gravitational force
F = kx = mg
x = 0.07, m = 0.1 kg, g = 9.81 m/s²
0.07k = 0.1 * 9.8
k = (0.1*9.8)/0.07
k = 14 N/m
When the force, F = 3N
F = kx
3 = 14x
x = 3/14
x = 0.214 m
The position of the block = 0.214 - 0.07 = 0.144m
B) Determine the work you did stretching the spring.
Energy stored in the spring when x = 0.07
E = 0.5 kx²
E = 0.5 * 14 * 0.07²
E = 0.0343 J
Energy stored in the spring when x = 0.214
E = 0.5 kx²
E = 0.5 * 14 * 0.214²
E = 0.32 J
Potential energy lost due to gravity = mgh
PE = 0.1 * 9.81 * 0.144
PE = 0.141 J
So to calculate the work done:
0.0343 + W = 0.32 - 0.141
W = 0.15 J
c) Energy in the spring
E = 0.32 - 0.0343 - 0.15
E = 0.1357 = 0.14 J
d)
[tex]1/2 *k *0.214^{2} = 1/2 kx^{2} + mg(0.214+x)\\0.32 = 7x^{2} + 0.1*9.8(0.214+x)\\[/tex]
Solving for x, x = 0.07 m
The block will rise 0.07m after it is released
A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made.
Answer:
(a) Current is 2831.93 A
(b) [tex]8.40A/m^2[/tex]
(c) [tex]\rho =15.52\times 10^{-9}ohm-m[/tex]
Explanation:
Length of wire l = 3.22 m
Diameter of wire d = 7.32 mm = 0.00732 m
Cross sectional area of wire
[tex]A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2[/tex]
Resistance [tex]R=11.9mohm=11.9\times 10^{-3}ohm[/tex]
Potential difference V = 33.7 volt
(A) current is equal to
[tex]i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A[/tex]
(B) Current density is equal to
[tex]J=\frac{i}{A}[/tex]
[tex]J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2[/tex]
(c) Resistance is equal to
[tex]R=\frac{\rho l}{A}[/tex]
[tex]11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}[/tex]
[tex]\rho =15.52\times 10^{-9}ohm-m[/tex]
A projectile with a mass of 0.100 kg is fired at and embeds itself in a stationary target with a mass of 2.57 kg. The target and embedded projectile fly off after being struck. Determine the percent of the projectile's incident kinetic energy carried off by the target-projectile combination.
Answer:
3.74 %
Explanation:
Given,
Mass of the fired projectile, m = 0.1 Kg
Mass of the stationary target, M = 2.57 Kg
Speed of the particle after collision
Using conservation of momentum
m v = (m + M) V
0.1 x v = (0.1 + 2.57) V
V = 0.0374 v
Initial KE = [tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}\times 0.1\times v^2[/tex]
Final KE = [tex]\dfrac{1}{2} (M + m) V^2 = \dfrac{1}{2}\times 2.67\times (0.0374 v)^2[/tex]
Now,
the percent of the projectile's KE carried out by target
[tex]=\dfrac{\dfrac{1}{2}\times 2.67\times (0.0374 v)^2}{ \dfrac{1}{2}\times 0.1\times v^2}\times 100[/tex]
[tex]= 3.74\ \%[/tex]
A simple pendulum is made my attaching a rod of negligible mass to a 2.0 kg pendulum bob at the end. It is observed that on Earth, the period of small-angle oscillations is 1.0 second. It is also observed that on Planet X this same pendulum has a period of 1.8 seconds. How much does the pendulum bob weigh on Planet X
Final answer:
The weight of the pendulum bob on Planet X remains 2.0 kg, the same as on Earth, because the mass of the pendulum bob does not affect the period of a simple pendulum. The difference in periods indicates a change in gravitational strength, not mass.
Explanation:
The question revolves around understanding how the period of oscillation of a simple pendulum changes with gravity on different planets. Since the mass of the pendulum bob does not affect the period of a simple pendulum, which depends only on the length of the pendulum and the gravitational acceleration (g), the weight of the pendulum bob on Planet X would still be 2.0 kg, the same as on Earth. However, the difference in periods between Earth and Planet X indicates a difference in gravitational acceleration, implying that g on Planet X is weaker than on Earth. The formula for the period (T) of a simple pendulum is T = 2π√(l/g), where l is the length of the pendulum and g is the gravitational acceleration. Since the mass of the pendulum bob does not factor into this equation, the weight of the pendulum bob on Planet X remains the same, but its apparent weight will change according to the planet's gravitational pull.
5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center B and angular velocity ω = 1 deg/s about a vertical axis. The velocity of B is constant, but the angular rate ω is decreasing at 0.5 deg/s2. Person A is stationary on the dock. What velocity and acceleration of A are observed by a passenger fixed to and rotating with the ship? Treat the problem as two‐dimensional.
The image is missing, so i have attached it;
Answer:
A) V_rel = [-(2.711)i - (0.2588)j] m/s
B) a_rel = (0.8637i + 0.0642j) m/s²
Explanation:
We are given;
the cruise ship velocity; V_b = 1 m/s
Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s
Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²
Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)
Thus,
V_a = V_b + (ω•r_ba) + V_rel
Now, V_a = 0. Thus;
0 = V_b + (ω•r_ba) + V_rel
V_rel = -V_b - (ω•r_ba)
From the image and plugging in relevant values, we have;
V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)
V_rel = - (cos15)i - (sin15)j - 1.745i
Note that; k x j = - i
V_rel = [-(2.711)i - (0.2588)j] m/s
B) Let's write the acceleration at A with respect to B in terms of a_b.
Thus,
a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel
a_a and a_b = 0.
Thus;
0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel
a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)
Plugging in the relevant values with their respective position vectors, we have;
a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])
a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i
Note that; k x j = - i and k x i = j
Thus,simplifying further ;
a_rel = 0.8637i + 0.0642j m/s²
Final answer:
From the ship's frame of reference, person A has a velocity of -1 m/s (opposite to the ship's motion). The acceleration observed would be the negative ship's angular deceleration converted to linear acceleration, needing more data for calculation.
Explanation:
The problem presents a rotational kinematics and relative motion scenario common in physics. Person A is stationary on the dock while the ship moves with velocity vB = 1 m/s and has an angular velocity of ω = 1 deg/s, decreasing at 0.5 deg/s2. From the ship's reference frame, the velocity of person A will appear as the opposite of the ship's velocity, which is -1 m/s. However, since person A is stationary, A's acceleration as observed from the ship would be the opposite of the ship's angular deceleration translated into linear acceleration at the radius person A is perceived to be from the ship's center of rotation, which requires further information to calculate.
A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kgm2 and a radius of 12.5 cm. A mass of 423 grams is attached to the free end of the string. With the string vertical and taut, the mass is gently released so it can descend under the influence of gravity. As the mass descends, the string unwinds and causes the pulley to rotate, but does not slip on the pulley. What is the speed (in m/s) of the mass after it has fallen through 1.25 m
Answer:
the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
Explanation:
Given that :
Mass attached to the free end of the string, m = 423 g = 0.423 kg
Moment of inertia of pulley, I = 0.0352 kg m²
Radius of the pulley, r = 12.5 cm = 0.125 meters
Depth of fallen mass, h = 1.25 m
Acceleration due to gravity, g = 9.8 m/s²
Change in potential energy = mgh
= 0.423 × 9.8 × 1.25
=5.18175 J
From the question, we understand that the change in potential energy is used to raise and increase the kinetic energy of hanging mass and the rotational kinetic energy of pulley.
As such;
[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2[/tex]
where;
[tex]\omega[/tex] is the angular velocity of the pulley
v is the velocity of the mass after falling 1.25 m
where:
[tex]v = r \omega[/tex]
[tex]\omega = \frac{v}{r}[/tex]
replacing [tex]\omega = \frac{v}{r}[/tex] into above equation; we have:
[tex]5.18175 \ J= \frac{1}{2}mv^2 + \frac{1}{2} I( \frac{v}{r})^2[/tex]
[tex]5.18175= (\frac{1}{2} m + \frac{1}{2}* (\frac{I}{r^2})v^2 \\ \\ v^2 = \frac{5.18175}{0.5*0.423+0.5*\frac{0.0352}{0.1252}} \\ \\ v^2 = 3.873047 \\ \\ v = 1.968 \ m/s[/tex]
Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s
SpeedWhat all information we have?
Mass attached to the free end of the string, m = 423 g = 0.423 kg
Moment of inertia of pulley, I = 0.0352 kg m²
Radius of the pulley, r = 12.5 cm = 0.125 meters
Depth of fallen mass, h = 1.25 m
Acceleration due to gravity, g = 9.8 m/s²
Change in potential energyChange in potential energy = mgh
Change in potential energy = 0.423 × 9.8 × 1.25
Change in potential energy =5.18175 J
As such:
5.18175 J= mv²+1w²
where;
w is the angular velocity of the pulley
v is the velocity of the mass after falling 1.25 m
v=rw
w=v/r
Replacing into above equation;
5.18175 J=mv² + 1/2 (w/r²) ²
5.18175 = (m + /* (4) v²
v² = 3.873047
v² =1.968 m/s
Thus, the speed (in m/s) of the mass after it has fallen through 1.25 m is 1.968 m/s.
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During a track and field event, a metal javelin (not infinitesimally thin) is thrown due east and parallel to the ground. At the field where the event took place, the magnetic field is running straight up and perpendicular to the ground. In which direction will the current flow in the metal object?
Answer:
There is no induced current
Explanation:
[Find the attachment]
Rotational dynamics about a fixed axis: A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular velocity of 2.40 rev/s, it stops after turning through 18.2 rev with uniform acceleration. The net torque acting on this sphere as it is slowing down is closest to:A) 0.149 N m. B) 0.0620N m. C) 0.00593 N m. D) 0.0372 N m. E) 0.0466 N·m
Answer:
D) 0.0372 N m
Explanation:
r = 45/2 cm = 22.5 cm = 0.225 m
As 1 revolution = 2π rad we can convert to radian unit
2.4 rev/s = 2.4 * 2π = 15.1 rad/s
18.2 rev = 18.2 * 2π = 114.35 rad
We can calculate the angular (de)acceleration using the following equation of motion
[tex]-\omega^2 = 2\alpha \theta [/tex]
[tex]- 15.1^2 = 2*\alpha * 114.35[/tex]
[tex]\alpha = \frac{-15.1^2}{2*114.35} = -0.994 rad/s^2[/tex]
The moment of inertia of the solid uniform sphere is
[tex]2mr^2/5 = 2*1.85*0.225^2/5 = 0.0375 kgm^2[/tex]
The net torque acting on this according to Newton's 2nd law is
[tex]T = I\alpha = 0.0375 * 0.994 = 0.0372 Nm[/tex]
Answer:
(D) The net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
Explanation:
Given;
mass of the solid sphere, m = 1.85 kg
radius of the sphere, r = ¹/₂ of diameter = 22.5 cm
initial angular velocity, ω = 2.40 rev/s = 15.08 rad/s
angular revolution, θ = 18.2 rev = 114.37 rad
Torque on the sphere, τ = Iα
Where;
I is moment of inertia
α is angular acceleration
Angular acceleration is calculated as;
[tex]\omega_f^2 = \omega_i^2 +2 \alpha \theta\\\\0 = 15.08^2 + (2*114.37)\alpha\\\\\alpha = \frac{-15.08^2}{(2*114.37)} = -0.994 \ rad/s^2\\\\\alpha = 0.994 \ rad/s^2 \ (in \ opposite \ direction)[/tex]
moment of inertia of solid sphere, I = ²/₅mr²
= ²/₅(1.85)(0.225)²
= 0.03746 kg.m²
Finally, the net torque on the sphere is calculated as;
τ = Iα
τ = 0.03746 x 0.994
τ = 0.0372 N.m
Therefore, the net torque acting on this sphere as it is slowing down is closest to 0.0372 N.m
According to the Heisenberg uncertainty principle, quantum mechanics differs from classical mechanics in that: Select the correct answer below: Quantum mechanics involves particles that do not move. It is impossible to calculate with accuracy both the position and momentum of particles in classical mechanics. The measurement of an observable quantity in the quantum domain inherently changes the value of that quantity. All of the above
Answer:
Statement 3 is correct.
Heisenberg's uncertainty principle explains that the measurement of an observable quantity in the quantum domain inherently changes the value of that quantity
Explanation:
Classical mechanics is the study of motion of big, relatable bodies that we come in contact with in our day to day lives.
Quantum mechanics refers to this same study, but for particles on a subatomic level.
Obviously, Classical mechanics' theories and principles were first discovered and they worked for their intended uses (still work!). But when studies on particles on a sub-atomic level intensified, it became impractical to apply those theories and principles to these sub-atomic particles that displayed wave-particle duality nature properly.
Heisenberg's Uncertainty principle came in a time that explanations and justifications were needed to adapt these theories to sub-atomic particles.
The principle explains properly that it is impossible to measure the position and velocity (momentum) of a sub-atomic particle in exact terms and at the same time.
Mathematically, it is presented as
Δx.Δp ≥ ℏ
Where ℏ= adjusted Planck's constant.
ℏ= (h/2π)
And Δx and Δp are the uncertainties in measuring the position and momentum of sub-atomic particles.
The major reason for this is the wave-particle duality of sub-atomic particles. They exist as waves and particles at the same time that a complete knowledge of their position mean that a complete ignorance of their velocity and vice versa.
Taking the statements one at a time
Statement 1
Quantum Mechanics studies sub-atomic particles which are mostly always in motion. So, this is false.
Statement 2
It is impossible to calculate with accuracy both the position and momentum of particles in quantum mechanics not classical mechanics. As stated above, the reason for the uncertainty is the wave-particle duality of sub-atomic particles which the particle in classical mechanics do not exhibit obviously enough.
Statement 3
Any attempt to measure precisely the velocity of a subatomic particle, will knock it about in an unpredictable way, so that a simultaneous measurement of its position has no validity.
An essential feature of quantum mechanics is that it is generally impossible, even in principle, to measure a system without disturbing it. This is basically the uncertainty principle rephrased. This is the only true statement.
Hope this Helps!!!
A solid cylinder with radius 0.160 m is mounted on a frictionless, stationary axle that lies along the cylinder axis. The cylinder is initially at rest. Then starting at t = 0 a constant horizontal force of 5.00 N is applied tangentially to the surface of the cylinder. You measure the angular displacement θ−θ0 of the cylinder as a function of the time t since the force was first applied. When you plot θ−θ0 (in radians) as a function of t2 (in s2), your data lie close to a straight line.
What is the moment of inertia of the wheel?
The moment of inertia of the solid cylinder can be found by calculating the torque and the angular acceleration of the cylinder, and then relating these quantities using the equation torque equals moment of inertia times angular acceleration.
Explanation:The moment of inertia of the solid cylinder can be found using the relationship between torque acting on the cylinder and its angular acceleration, both of which can be inferred from the given information. The torque τ on the cylinder is given by the product of the force applied and the radius of the cylinder, τ = Fr. Since this force is causing the cylinder to undergo rotational motion, it is also causing an angular acceleration α, which can be calculated from the slope of plot relating angular displacement θ and time t squared, as α is directly proportional to slope of θ vs t² plot.
By relating these variables using the equation τ = Iα, where I is the moment of inertia, we can solve for I as: I = τ/α = Fr/α. By substituting given values, we can compute specific value for moment of inertia.
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The metal zinc has free electron concentration per unit volume, nv, of 1.31029/m3 and an electron mobility me of 8104 m2/V.s. The charge carried by an electron, e, is 1.61019 coulomb. Based on this information, what is the electrical conductivity of zinc? Handbooks list the measured resistivity of zinc as 5.9 mU.cm. Is this consistent with your calculation? (Watch the units.)
Answer:
1.7*10^{12}Ucm^-1
Explanation:
The answer to this question is obtained by using the following formula:
[tex]\sigma=nq\mu_e[/tex]
sigma: conductivity
q: charge of the electron = 1.61019*10^{-19}C
mu_e: electron mobility = 8104 m^2/Vs
n: free electron concentration = 1.31029/m^3
By replacing you get:
[tex]\sigma=(1.31*10^{29}/m^3)(1.61*10^{-19}C)(8104 m2/Vs)=1.709*10^{14}\Omega^{-1}m^{-1} \\\\1.709*10^{14}\Omega^{-1} (1*10^{2}cm})^{-1}=1.7*10^{12}\Omega^{-1}cm^{-1}[/tex]
the result obained is not 5.9mU.cm. This is because temperature effects has not taken into account.
The resistance of physiological tissues is quite variable. The resistance of the internal tissues of humans, primarily composed of salty solutions, is quite low. Here the resistance between two internal points in the body is on the order of 100 ohms. Dry skin, however, can have a very high resistance, with values ranging from thousands to hundreds of thousands of ohms. However, if skin is wet, it is far more conductive, and so even contact with small voltages can create large, dangerous currents though a human body. (For example, although there is no specific minimum current that is lethal, currents generally exceeding a couple tenths of Amps may be deadly.)
Assuming that electrocution can be prevented if currents are kept below 0.1 A, and assuming the resistance of dry skin is 100,000 ohms, what is the maximum voltage with which a person could come into contact while avoiding electrocution? (Of course, all bets are off and things become far more dangerous if this person's skin is wet, which can reduce the resistance by more than a factor of 100.)
Given Information:
Current = I = 0.1 A
Resistance = R = 100 kΩ
Required Information:
Voltage = V = ?
Answer:
Voltage = V = 1000 V
Step-by-step explanation:
We know that electrocution depends upon the amount of current flowing through the body and the voltage across the body.
V = IR
Where I is the current flowing through the body and R is the resistance of body.
If electrocution can be avoided when the current is below 0.1 A then
V = 0.1*10×10³
V = 1000 Volts
Therefore, 1000 V is the maximum voltage with which a person could come into contact while avoiding electrocution, any voltage more than 1000 V may result in fatal electrocution.
Also note that human body has very low resistance when the body is wet therefore, above calculated value would not be applicable in such case.
An astronaut working with many tools some distance away from a spacecraft is stranded when the "maneuvering unit" malfunctions. How can the astronaut return to the spacecraft by sacrificing some of the tools? (Note: the maneuvering unit is connected to the astronaut's spacesuit and is not removable)
Answer:
He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.
Explanation:
By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.
An astronaut stranded far from their spacecraft can return by using the principle of Newton's third law of motion. By throwing tools in the opposite direction of the spacecraft, the astronaut can create a reaction force that propels them back to the spacecraft.
Explanation:In this scenario, the astronaut working with many tools some distance away from their spacecraft can use the concept of Newton's third law of motion, also known as action-reaction principle, to return to the spacecraft. According to this law, for every action, there is an equal and opposite reaction. So, the astronaut can return to the spacecraft by sacrificing some of the tools that they are carrying, and throwing the tools in the opposite direction of the spacecraft. When the tools are thrown away, the reaction force will propel the astronaut towards the spacecraft. This is a effective method of maneuvering in space because in the vacuum of space, even small forces can have a significant impact due to the absence of air resistance or friction. This way, the astronaut can cleverly convert a threat into an opportunity and solve their problem with minimal risks.
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Ideal incompressible water is flowing in a drainage channel of rectangular cross-section. At one point, the width of the channel is 12 m, the depth of water is 6.0 m, and the speed of the flow is 2.5 m/s. At a point downstream, the width has narrowed to 9.0 m, and the depth of water is 8.0 m. What is the speed of the flow at the second point?a. 3.0 m/s b. 2.0 m/s c. 4.0 m/s d. 2.5 m/s e. 3.5 m/s
Answer:
(D) The speed of the flow at the second point is 2.5 m/s
Explanation:
Given;
at upstream;
the width of the channel, w = 12 m
the depth of water, d = 6.0 m
the speed of the flow, v = 2.5 m/s
at downstream;
the width of the channel, w = 9 m
the depth of water, d = 8.0 m
the speed of the flow, v = ?
Volumetric flow rate for an Ideal incompressible water is given as;
Q = Av
where;
A is the area of the channel of flow
v is velocity of flow
For a constant volumetric flow rate;
A₁v₁ = A₂v₂
A₁ = area of rectangle = L x d = 12 x 6 = 72 m²
A₂ = area of rectangle = L x d = 9 x 8 = 72 m²
(72 x 2.5) = 72v₂
v₂ = 2.5 m/s
Therefore, the speed of the flow at the second point is 2.5 m/s
A light source shines light consisting of two wavelengths, λ1 = 540 nm (green) and λ2 = 450 nm (blue), on two slits separated by 0.180 mm. The two overlapping interference patterns, one from each wavelength, are observed on a screen 1.53 m from the slits. What is the minimum distance (in cm) from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light?
Answer:
The maximun distance is [tex]z_1 = z_2 = 0.0138m[/tex]
Explanation:
From the question we are told that
The wavelength are [tex]\lambda _ 1 = 540nm (green) = 540 *10^{-9}m[/tex]
[tex]\lambda_2 = 450nm(blue) = 450 *10^{-9}m[/tex]
The distance of seperation of the two slit is [tex]d = 0.180mm = 0.180 *10^{-3}m[/tex]
The distance from the screen is [tex]D = 1.53m[/tex]
Generally the distance of the bright fringe to the center of the screen is mathematically represented as
[tex]z = \frac{m \lambda D}{d}[/tex]
Where m is the order of the fringe
For the first wavelength we have
[tex]z_1 = \frac{m_1 (549 *10^{-9} * (1.53))}{0.180*10^{-3}}[/tex]
[tex]z_1=0.00459m_1 m[/tex]
[tex]z_1= 4.6*10^{-3}m_1 m ----(1)[/tex]
For the second wavelength we have
[tex]z_2 = m_2 \frac{450*10^{-9} * 1.53 }{0.180*10^{-3}}[/tex]
[tex]z_2 = 0.003825m_2[/tex]
[tex]z_2 = 3.825 *10^{-3} m_2 m[/tex] ----(2)
From the question we are told that the two sides coincides with one another so
[tex]zy_1 =z_2[/tex]
[tex]4.6*10^{-3}m_1 m = 3.825 *10^{-3} m_2 m[/tex]
[tex]\frac{m_1}{m_2} = \frac{3.825 *10^{-3}}{4.6*10^{-3}}[/tex]
Hence for this equation to be solved
[tex]m_1 = 3[/tex]
and [tex]m_2 = 4[/tex]
Substituting this into the equation
[tex]z_1 = z_2 = 3 * 4.6*10^{-3} = 4* 3.825*10^{-3}[/tex]
Hence [tex]z_1 = z_2 = 0.0138m[/tex]
The minimum distance from the center of the screen where green and blue bright fringes coincide is calculated using principles of double-slit interference. Applying these principles and trigonometry, we find the minimum distance from the center of the screen to this point.
Explanation:To find the minimum distance from the center of the screen to a point where a bright fringe of the green light coincides with a bright fringe of the blue light, we can use the formula for double slit interference:
d sin θ = mλ
where d is the separation of the slits, θ is the angle from the central bright fringe, m is the order of fringe, and λ is the wavelength of light.
For the green and blue lights to interfere constructively on the screen for the first time, their path length difference must be equal to their respective wavelengths. Therefore, the order of fringe for the green light will be different from the blue light. The minimum distance can be found when the fringe order for the green light (m1) is 1 and for the blue light (m2) is, by proportion of the wavelengths, rounded to the nearest whole number. Solving the equation for each colour and setting them equal will give the minimum distance.
Finally, the distance from the center to the fringe can be calculated using the trigonometric relationship:
L tan θ, where L is the distance from the slits to the screen.
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Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point.
Answer:
See attached handwritten document for answer
Explanation:
Answer:
Explanation:
a) You can compute the force by using the expression:
[tex]F=k\frac{q_1q_2}{[(x-x_o)^2+(y-y_o)^2+(z-z_o)^2]^{\frac{1}{2}}}[/tex]
where k=8.98*10^9Nm^2/C^2 and q1, q2 are the charges. By replacing for the forces you obtain:
[tex]F_T=k[\frac{(1*10^{-3}C)(10*10^{-9}C)}{[(3-0)^2+(2-3)^2+(-1-1)^2]}}][(3-0)\hat{i}+(2-3)\hat{j}+(-1-1)\hat{k}]\\\\ \ \ \ \ +k[\frac{(-2*10^{-3}C)(10*10^{-9}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}}][(-1-0)\hat{i}+(-1-3)\hat{j}+(4-1)\hat{k}]\\\\F_T=6.41*10^{-3}N[3i-j-2k]-6.9*10^{-3}N[-1i-4j+3k]\\\\=0.026N\hat{i}-0.034N\hat{j}-0.033\hat{k}[/tex]
b)
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}]+k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}]\\\\E=641428.5N/C+690769.23N/C=1332197.73N/C[/tex]
[tex]E=k[\frac{1*10^{-3}C}{[(3-0)^2+(2-3)^2+(-1-1)^2]}][3i-2j-2k]+\\\\k[\frac{(-2*10^{-3}C)}{[(-1-0)^2+(-1-3)^2+(4-1)^2]}][-1i-4j+3k]\\\\E=641428.5N/C[3i-2j-2k]-690769.23N/C[-1i-4j+3k]=2615054.7i+1480219.92j-4973626.23k\\\\|E|=\sqrt{(E_x)^2+(E_y)^2+(E_z)^2}=5810896.56N/C[/tex]
where we you have used that E=kq/r^2
A 1.0-kg standard cart A collides with a 0.10-kg cart B. The x component of the velocity of cart A is +0.70 m/s before the collision and +0.50 m/s after the collision. Cart B is initially traveling toward cart A at 0.40 m/s , and after the collision the x component of its velocity is +1.6 m/s .
Part A: What is the x component of the change in the momentum of A?
Part B: What is the x component of the change in the momentum of B?
Part C: What is the sum of these two x components of the changes in momentum?
The x component of the change in momentum of cart A is -0.20 kg·m/s, while for cart B, it is +0.20 kg·m/s. The sum of these changes in momentum is 0 kg·m/s, showing conservation of momentum in the system.
Explanation:The student has provided information about a collision between two carts, which involves concepts of momentum and conservation of momentum from mechanics in physics.
Part A: Change in Momentum of Cart A
The x component of the change in momentum of cart A can be calculated using the formula Δp = m(v_final - v_initial), where m is mass and v is velocity. Here, Δp = 1.0 kg (0.50 m/s - 0.70 m/s) = -0.20 kg·m/s. The negative sign indicates that the momentum decreased.
Part B: Change in Momentum of Cart B
Similarly, for cart B, Δp = 0.10 kg (1.6 m/s - (-0.40 m/s)) = 0.20 kg·m/s. The positive sign signifies an increase in momentum.
Part C: Sum of Changes in Momentum
The total change in momentum for both carts is the sum of their individual momentum changes, which is -0.20 kg·m/s + 0.20 kg·m/s = 0 kg·m/s. This means that the total momentum of the system is conserved.
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 60.0 cm long is at a location 4.00 mi from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.
To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is
[tex]I = \frac{P_{avg}}{A}[/tex]
[tex]I = \frac{P_{avg}}{4\pi r^2}[/tex]
[tex]I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m[/tex]
[tex]I = 6.529*10^{-6}W/m^2[/tex]
The amplitude of electric field at the receiver is
[tex]I = \frac{E_{max}^2}{2\mu_0 c}[/tex]
[tex]E_{max}= \sqrt{2I\mu_0 c}[/tex]
The amplitude of induced emf by this signal between the ends of the receiving antenna is
[tex]\epsilon_{max} = E_{max} d[/tex]
[tex]\epsilon_{max} = \sqrt{2I \mu_0 cd}[/tex]
Here,
I = Current
[tex]\mu_0[/tex] = Permeability at free space
c = Light speed
d = Distance
Replacing,
[tex]\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}[/tex]
[tex]\epsilon_{max} = 0.05434V[/tex]
Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity ( Δ v ) Δ x ≥ h 4 π m Δ v where h is Planck's constant and m is the mass of the object. The mass of an electron is 9.11 × 10 − 31 kg. What is the uncertainty in the position of an electron moving at 2.00 × 10 6 m/s with an uncertainty of Δ v = 0.01 × 10 6 m/s ?
According to the information given, the Heisenberg uncertainty principle would be given by the relationship
[tex]\Delta x \Delta v \geq \frac{h}{4\pi m}[/tex]
Here,
h = Planck's constant
[tex]\Delta v[/tex] = Uncertainty in velocity of object
[tex]\Delta x[/tex] = Uncertainty in position of object
m = Mass of object
Rearranging to find the position
[tex]\Delta x \geq \frac{h}{4\pi m\Delta v}[/tex]
Replacing with our values we have,
[tex]\Delta x \geq \frac{6.625*10^{-34}m^2\cdot kg/s}{4\pi (9.1*10^{-31}kg)(0.01*10^6m/s)}[/tex]
[tex]\Delta x \geq 5.79*10^{-9}m[/tex]
Therefore the uncertainty in position of electron is [tex]5.79*10^{-9}m[/tex]
A student builds and calibrates an accelerometer and uses it to determine the speed of her car around a certain unbanked highway curve. The accelerometer is a plumb bob with a protractor that she attaches to the roof of her car. A friend riding in the car with the student observes that the plumb bob hangs at an angle of 15.0° from the vertical when the car has a speed of 21.5 m/s. (a) What is the centripetal acceleration of the car rounding the curve? m/s2 (b) What is the radius of the curve? m (c) What is the speed of the car if the plumb bob deflection is 7.00° while rounding the same curve? m/s
(a) [tex]\(a_c \approx 39.8 \, \text{m/s}^2\)[/tex]
(b) [tex]\(r \approx 11.6 \, \text{m}\)[/tex]
(c) [tex]\(v_2 \approx 28.4 \, \text{m/s}\)[/tex]
(a) calculation of the centripetal acceleration [tex](\(a_c\))[/tex],
[tex]\[ a_c = \frac{g}{\sin(\theta_1)} \][/tex]
Given [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_1 = 15.0°\)[/tex], we find:
[tex]\[ a_c = \frac{9.8 \, \text{m/s}^2}{\sin(15.0°)} \][/tex]
Calculating this gives [tex]\(a_c \approx 39.8 \, \text{m/s}^2\).[/tex]
(b) The radius ((r)) of the curve can be found using the formula:
[tex]\[ r = \frac{v^2}{a_c} \][/tex]
Substituting [tex]\(v = 21.5 \, \text{m/s}\) and \(a_c \approx 39.8 \, \text{m/s}^2\),[/tex]we get:
[tex]\[ r = \frac{(21.5 \, \text{m/s})^2}{39.8 \, \text{m/s}^2} \][/tex]
This calculation yields [tex]\(r \approx 11.6 \, \text{m}\).[/tex]
(c) For a new deflection angle [tex]\(\theta_2 = 7.00°\)[/tex], we find the new centripetal acceleration [tex](\(a_{c2}\))[/tex] using:
[tex]\[ a_{c2} = \frac{g}{\sin(\theta_2)} \][/tex]
Substituting [tex]\(g \approx 9.8 \, \text{m/s}^2\) and \(\theta_2 = 7.00°\)[/tex], we get:
[tex]\[ a_{c2} = \frac{9.8 \, \text{m/s}^2}{\sin(7.00°)} \][/tex]
Calculating this gives [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\).[/tex]
Then, using [tex]\(a_{c2} = \frac{v_2^2}{r}\)[/tex], we find the new speed [tex](\(v_2\)):[/tex]
[tex]\[ v_2 = \sqrt{a_{c2} \cdot r} \][/tex]
Substituting [tex]\(a_{c2} \approx 84.7 \, \text{m/s}^2\) and \(r \approx 11.6 \, \text{m}\), we get \(v_2 \approx 28.4 \, \text{m/s}\).[/tex]