Answer:
Probably all but (a)
The Tyndall Effect is caused by dispersion of the incident light by the individual molecules in the liquid.
Salt water and a foggy night will cause dispersion of the incident light.
A soda drink may also cause this dispersion, but and not sure.
If Vector A is (6, 4) and Vector B is (-2, -1), what is A – B?
A. (8,5)
B. (4,5)
C. (4,3)
D. (8,3)
Answer:
I think the answer is A...I'm not sure
Explanation:
A=(6,4)
B=(-2,-1)
A-B=(6-(-2)),(4-(-1))
=(6+2),(4+1)
=(8,5)
Answer:
[tex]6-(-2)=[/tex]
[tex]6+2[/tex]
[tex]=8[/tex]
[tex]4-\left(-1\right)[/tex]
[tex]=4+1[/tex]
[tex]=5[/tex]
[tex](8,5)[/tex]
[tex]\textbf{OAmalOHopeO}[/tex]
Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
The element sodium can emit light at two wavelengths, λ1 = 588.9950 nm and λ2 = 589.5924 nm. Light in sodium is being used in a Michelson interferometer. Through what distance must mirror M 2 be moved if the shift in the fringe pattern for one wavelength is to be 1.00 fringe more than the shift in the fringe pattern for the other wavelength?
Answer:
The distance is [tex]d = 0.00029065 \ m[/tex]
Explanation:
From the question we are told that
The first wavelength is [tex]\lambda _1 = 588.9950 nm = 588.9950 *10^{-9} \ m[/tex]
The second wavelength is [tex]\lambda _2 = 589.5924 nm = 589.5924 *10^{-9} \ m[/tex]
The difference in the fringe pattern is n = 1.0
Generally the equation defining the effect of the movement of the mirror M 2 in a Michelson interferometer is mathematically represented as
[tex]2 * d = [\frac{\lambda _1 * \lambda_2 }{\lambda_2 - \lambda _1 } ] * n[/tex]
Here d is the mirror M 2 must be moved
substituting values
[tex]2 * d = [\frac{(588.9950*10^{-9} ) * (589.5924 *10^{-9}) }{(589.5924 *10^{-9}) - (588.9950*10^{-9} ) } ] * 1.0[/tex]
[tex]d = 0.00029065 \ m[/tex]
find the velocity of the object for all relevent times find the position of the object for all relevent times a softball is popped up vertically velocity of 32 m/s
Answer:
whats the formula
Explanation:
In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in s) of the material
Answer:
2113 secondsExplanation:
The general decay equation is given as [tex]N = N_0e^{-\lambda t} \\\\[/tex], then;
[tex]\dfrac{N}{N_0} = e^{-\lambda t} \\[/tex] where;
[tex]N/N_0[/tex] is the fraction of the radioactive substance present = 1/16
[tex]\lambda[/tex] is the decay constant
t is the time taken for decay to occur = 8,450s
Before we can find the half life of the material, we need to get the decay constant first.
Substituting the given values into the formula above, we will have;
[tex]\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328[/tex]
Half life f the material is expressed as [tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]
[tex]t_{1/2} = \frac{0.693}{0.000328}[/tex]
[tex]t_{1/2} = 2,112.8 secs[/tex]
Hence, the half life of the material is approximately 2113 seconds
In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.
Answer:
1.24Mev
Explanation:
Using
E= hc/lambda
= (6.62x10^-19) x(3x10^8m/s)/(1x10^-12) x 1.602x10^-9
= 1.24Mev
In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force
Answer:
E = VdB
Explanation:
This is because canceling the electric and magnetic force means
q.vd. B= we
E= Vd. B
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
What do we call a substance in
which two or more elements are
chemically bonded
Answer:
A compound
Explanation:
A compound is a substance formed when two or more elements are chemically joined
Answer:
Compound
Explanation:
A compound is a substance derived from the chemical combination of two or more elements
e.g Water ;
= [tex]H_2O\\Hydrogen\:and\:Oxygen[/tex]
Salt ;
[tex]NaCl\\Sodium\:and\: Chlorine[/tex]
A bar magnet is dropped from above and falls through the loop of wire. The north pole of the bar magnet points downward towards the page as it falls. Which statement is correct?a. The current in the loop always flows in a clockwise direction. b·The current in the loop always flows in a counterclockwise direction. c. The current in the loop flows first in a clockwise, then in a counterclockwise direction. d. The current in the loop flows first in a counterclockwise, then in a clockwise direction. e. No current flows in the loop because both ends of the magnet move through the loop.
Answer:
b. The current in the loop always flows in a counterclockwise direction.
Explanation:
When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.
The current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
The given problem is based on the concept and fundamentals of magnetic bars. When a magnet falls through a loop of wire, it induces an induced current on the loop of wire. There is some magnitude of current induced in the wire.
This induced current is due to the motion of the magnet through the loop, which cause a change in the flux linkage of the magnet. According to Lenz law, the induced current acts in such a way as to repel the force or action that produces it. For this magnet, the only opposition possible is to stop its fall by inducing a like pole on the wire loop to repel its motion down. An induced current that flows counterclockwise in the wire loop has a polarity that is equivalent to a north pole on a magnet, and this will try to repel the motion of the magnet through the coil. Also, when the magnet goes pass the wire loop, this induced north pole will try to attract the south end of the magnet, all in a bid to stop its motion downwards.Thus, we can say that the current in the loop always flows in a counterclockwise direction. Hence, option (b) is correct.
Learn more about the magnetic field here:
https://brainly.com/question/14848188
When an LRC series circuit is driven at resonance, which of the following statements about the circuit are correct? (There may be more than one correct choice.)a) The impedance of the circuit has its minimum value.b) The inductive reactance and the capacitive reactance are exactly equal to each other.c) The impedance of the circuit is zero.d) The inductive reactance and the capacitive reactance are both zero.e) The impedance of the circuit has its maximum value.
Answer:
Answers a) and b) should be marked as correct.
Explanation:
Recall that the resonance in an LRC circuit occurs when the current through the circuit is at its maximum, and such takes place when the impedance (Z) of the circuit reaches its maximum. This means that the impedance (see formula below) is at its minimum value:
[tex]Z=\sqrt{R^2+(\omega\,L-\frac{1}{\omega\,C})^2 }[/tex]
as per the impedance expression above, such happens when the term in parenthesis inside the root which contains the inductive reactance ([tex]\omega\,L[/tex]) and the capacitive reactance ([tex]1/\omega\,C[/tex]) have the same value.
Therefore, answers:
a) "The impedance of the circuit has its minimum value."
and
b) "The inductive reactance and the capacitive reactance are exactly equal to each other."
are correct answers.
(a) The impedance of the circuit has its minimum value.
(b) The inductive reactance and the capacitive reactance are exactly equal to each other
LRC series circuit consists of inductor, resistor and capacitor is series.
The impedance of the circuit is calculated as follows;
[tex]Z = \sqrt{R^2 + (X_C -X_L)^2}[/tex]
where;
R is the resistance[tex]X_C[/tex] is the capacitive reactance[tex]X_L[/tex] is the inductive reactanceThe impedance of the circuit is minimum when the capacitive reactance is equal to the inductive reactance.
[tex]X_C = X_L \\\\Z = \sqrt{R^2 \ + (0)^2} \\\\Z = R[/tex]
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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, and 130 cm. The loop is in a uniform magnetic field of magnitude 75.0 mT whose direction is parallel to the current in the 130 cm side of the loop. What is the magnitude of the magnetic force on the
Given that,
Current = 4 A
Sides of triangle = 50.0 cm, 120 cm and 130 cm
Magnetic field = 75.0 mT
Distance = 130 cm
We need to calculate the angle α
Using cosine law
[tex]120^2=130^2+50^2-2\times130\times50\cos\alpha[/tex]
[tex]\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}[/tex]
[tex]\alpha=\cos^{-1}(0.3846)[/tex]
[tex]\alpha=67.38^{\circ}[/tex]
We need to calculate the angle β
Using cosine law
[tex]50^2=130^2+120^2-2\times130\times120\cos\beta[/tex]
[tex]\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}[/tex]
[tex]\beta=\cos^{-1}(0.923)[/tex]
[tex]\beta=22.63^{\circ}[/tex]
We need to calculate the force on 130 cm side
Using formula of force
[tex]F_{130}=ILB\sin\theta[/tex]
[tex]F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0[/tex]
[tex]F_{130}=0[/tex]
We need to calculate the force on 120 cm side
Using formula of force
[tex]F_{120}=ILB\sin\beta[/tex]
[tex]F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63[/tex]
[tex]F_{120}=0.1385\ N[/tex]
The direction of force is out of page.
We need to calculate the force on 50 cm side
Using formula of force
[tex]F_{50}=ILB\sin\alpha[/tex]
[tex]F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38[/tex]
[tex]F_{50}=0.1385\ N[/tex]
The direction of force is into page.
Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.
a. The magnitude of the magnetic force on the 130 cm side is 0 Newton.
b. The magnitude of the magnetic force on the 120 cm side is 0.1385 Newton.
c. The magnitude of the magnetic force on the 50 cm side is 0.1385 Newton.
Given the following data:
Current = 4.00 Amperes.Magnetic field strength = 75.0 mT = [tex]7.5 \times 20^{-3}\;T[/tex]Length = 130 cm to m = 1.3 mHypotenuse = 130 cmOpposite side = 120 cmAdjacent side = 50 cmLet us assume the current is flowing in a counterclockwise direction in the right-angle triangle.
First of all, we would determine the angles by using cosine rule:
[tex]C^2=A^2 +B^2 - 2ABCos\alpha \\\\120^2=130^2 +50^2 - 2(130)(50)Cos\alpha\\\\14400 = 16900 + 2500 -13000Cos\alpha\\\\13000Cos\alpha=19400-14400 \\\\Cos\alpha=\frac{5000}{13000} \\\\\alpha = Cos^{-1}(0.3846)\\\\\alpha =67.38^\circ[/tex]
[tex]C^2=A^2 +B^2 - 2ABCos\beta \\\\50^2=120^2 +130^2 - 2(120)(130)Cos\beta \\\\2500 = 14400 + 16900 -31200Cos\beta\\\\31200Cos\alpha=31300-2500 \\\\Cos\beta=\frac{28800}{31200} \\\\\beta = Cos^{-1}(0.9231)\\\\\beta =22.62^\circ[/tex]
a. To the determine the magnitude of the magnetic force on the 130 cm side:
Mathematically, the force acting on a current in a magnetic field is given by the formula:
[tex]F = BILsin\theta[/tex]
Where:
B is the magnetic field strength.I is the current flowing through a conductor.L is the length of conductor.[tex]\theta[/tex] is the angle between a conductor and the magnetic field.Substituting the given parameters into the formula, we have;
[tex]F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times sin(0)\\\\F_{130}=7.5 \times 20^{-3}\times 4 \times 1.3 \times0\\\\F_{130}=0\;Newton[/tex]
b. To the determine the magnitude of the magnetic force on the 120 cm side:
[tex]F_{120}=BILsin\beta[/tex]
[tex]F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times sin(22.62)\\\\F_{120}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.3846\\\\F_{120}=0.1385\;Newton[/tex]
c. To the determine the magnitude of the magnetic force on the 50 cm side:
[tex]F_{50}=BILsin\alpha[/tex]
[tex]F_{50}=7.5 \times 20^{-3}\times 4 \times 0.5 \times sin(67.38)\\\\F_{50}=7.5 \times 20^{-3}\times 4 \times 1.2 \times0.9231\\\\F_{50}=0.1385\;Newton[/tex]
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Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion
Answer:
The distance is [tex]D = 0.000712 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light source is [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]
The distance from a pin hole is [tex]x = 9\ m[/tex]
The diameter of the pin hole is [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]
Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is
mathematically represented as
[tex]D = \frac{1.22 \lambda }{d }[/tex]
substituting values
[tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]
[tex]D = 0.000712 \ m[/tex]
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
An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA. Part A What is the resistance RR of the inductor
i
CHECK COMPLETE QUESTION BELOW
inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.
Part A)What is the resistance RR of the inductor
PART B) what is inductance L of the conductor
Answer:
A)R=1818.18 ohms
B)L=1.0446H
Explanation:
We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.
There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,
.
a)A)What is the resistance RR of the inductor?
The current flowing into RL circuit can be calculated using below expresion
i=ε/R[1-e⁻(R/L)t]
at t=∞ there is maximum current
i(max)= ε/R
Where ε emf of the battery
R is the resistance
R=ε/i(max)
= 12V/(6.60*10⁻³A)
R=1818.18 ohms
Therefore, the resistance R=1818.18 ohms
b)what is inductance L of the conductor?
i(t=0.80ms and 4.96mA
RT/L = ⁻ln[1- 1/t(max)]
Making L subject of formula we have
L=-RT/ln[1-i/i(max)]
If we substitute the values into the above expresion we have
L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]
L=1.0446H
Therefore, the inductor L=1.0446H
The double bond between two oxygen atoms (a molecule of oxygen air) has
two characteristics. What are they?
A. Four valence electrons are shared.
B. A metallic bond is formed.
C. Valence electrons are shared between oxygen atoms.
D. An ionic bond is formed.
Answer:
valance electrons are shared between oxygen atoms.. making them have eight in the outer most shells.
I hope this helps
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike the ground
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?
Answer:
Time taken for 1 swing = 3.81 second
Explanation:
Given:
Time taken for 1 swing = 2.20 Sec
Find:
Time taken for 1 swing , when triple the length(T2)
Computation:
Time taken for 1 swing = 2π[√l/g]
2.20 = 2π[√l/g].......Eq1
Time taken for 1 swing , when triple the length (3L)
Time taken for 1 swing = 2π[√3l/g].......Eq2
Squaring and dividing the eq(1) by (2)
4.84 / T2² = 1 / 3
T2 = 3.81 second
Time taken for 1 swing = 3.81 second
16. If one body is positively charged and another body is negatively charged, free electrons tend to
O A. move from the negatively charged body to the positively charged body
O B. remain in the positively charged body
OC. move from the positively charged body to the negatively charged body
O D. remain in the negatively charged body
Answer:
Hey there!
The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body
Let me know if this helps :)
A collector that has better efficiency in cold weather is the:
flat-plate collector due to reduced heat loss
evacuated tube collector due to its larger size
flat-plate collector due to the dark-colored coating
O evacuated tube collector due to reduced heat loss
Question 23 (1 point) Saved
One of the following is not found in Thermosyphon systems
o
Answer:
D. evacuated tube collector due to reduced heat loss
Explanation:
Evacuated tube collectors has vacuum which reduces the loss of heat and increase the efficiency of the collector. It has a major application in solar collector, and converts solar energy to heat energy. It can also be used for heating of a definite volume of water majorly for domestic purpose.
During cold weather, the conservation and efficient use of heat is required. Therefore, evacuated tube collector is preferred so as to reduce heat loss and ensure the maximum use of heat energy.
Mars Rover When the Mars rover was deployed on the surface of Mars in July 1997, radio signals took about 12 minmin to travel from Earth to the rover.
How far was Mars from Earth at that time?
Answer:
s = 2.16 x 10¹¹ m
Explanation:
Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:
s = vt
where,
s = the distance between Earth and Mars = ?
v = speed of the wave = speed of light = 3 x 10⁸ m/s
t = time taken by the radio signals to reach the rover from Earth
t = (12 min)(60 s/1 min) = 720 s
Therefore,
s = (3 x 10⁸ m/s)(720 s)
s = 2.16 x 10¹¹ m
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
The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. Which equations will you use to find n and v?
Answer:
n= speed of light in vacuum/ speed of light in the other medium.
Explanation:
If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;
n= speed of light in vacuum/ speed of light in the other medium.
Hence;
n= c/v
Where;
n= refractive index of the material
c= speed of light in vacuum
v = speed of light in another medium.
Note that the refractive index is the amount by which a transparent medium decreases the speed of light.
I REALLY NEED HELP WITH PHYSICS ASAP!!!
Vf^2 = v0^2 + 2a (xf - x0)
Solve for a
Answer:
a. solve for a
[tex]vf ^{2} = vo ^{2} + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2} - vo ^{2} \\ a = \frac{vf^{2} - vo^{2} }{2(xf - xo)} \\ a = \frac{vf ^{2} - vo ^{2} }{2xf - 2xo} [/tex]
I hope I helped you ^_^
Trong thực nghiệm nhiễu xạ qua 1 khe hẹp, bề rộng khe là 2.10-5 m, màn quan sát đặt cách khe 0,5 m. Chiếu ánh sáng có bước sóng 480 nm xuyên qua khe và thấy có nền nhiễu xạ trên màn quan sát. Tính bề rộng của vân sáng liền kề vân sáng trung tâm.
Answer:
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Explanation:
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A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an angle of 30 degrees with the horizontal. The worker believes that he can get the crate to the very top of the ramp by launching it at 5 m/s at the bottom and letting go. But friction is not neglible; the crate slides 1.6 m upthe ramp, stops, and slides back down.
Required:
a. Assuming that the friction force actingon the crate is constant, find its magnitude.
b. How fast is teh crate moving when it reachesthe bottom of the ramp?
Answer:
a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.
Explanation:
a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.
By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:
[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
Where:
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.
The equation is now simplified and the coefficient of friction is consequently cleared:
[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]
[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]
The final height of the crate is:
[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]
[tex]y_{2} = 0.8\,m[/tex]
If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:
[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\mu_{k} \approx 0.548[/tex]
Then, the magnitude of the friction force is:
[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]
If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:
[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]
[tex]f = 55.851\,N[/tex]
The magnitude of the force of friction is 55.851 newtons.
b) The energy equation of the situation is:
[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]
[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]
Now, the final speed is cleared:
[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]
[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]
Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:
[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]
The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.
The isotope (_90^234)Th has a half-life of 24days and decays to (_91^234)Pa. How long does it take for 90% of a sample of (_90^234)Th to decay to (_91^234)Pa?
Answer:
79.7 days
Explanation:
Half-life equation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is the amount of time,
and T is the half life.
If 90% decays, then 10% is left.
A = A₀ (½)^(t / T)
0.1 A₀ = A₀ (½)^(t / 24)
0.1 = ½^(t / 24)
ln(0.1) = (t / 24) ln(0.5)
t ≈ 79.7 days
how can you relazie a perfect balck body in pratice
Two different galvanometers G1 and G2, have internal resistances r1and r2. The galvanometers G1 and G2 require the same current IC1=IC2 for a full-scale deflection of their pointers. These galvanometers G1 and G2 are used to build lab-made ammeters A1 and A2 . Both ammeters A1 and A2 have the same maximum scale reading Imax1=Imax2=Imax. To build A1 ,shunt resistor of resistance Rsh1is used and to build A2 , shunt resistor of resistance Rsh2 is used. The value of these shunt resistor resistances are such that: Rsh1=3Rsh2. What is the ratio oftheir internal resistances: r1:r2?
Answer:
there are 3 photos attached. so check
Explanation: