Answer:
Approximately [tex]28^{\circ}[/tex].
Explanation:
The refractive index of the air [tex]n_{\text{air}}[/tex] is approximately [tex]1.00[/tex].
Let [tex]n_\text{glass}[/tex] denote the refractive index of the glass block, and let [tex]\theta _{\text{glass}}[/tex] denote the angle of refraction in the glass. Let [tex]\theta_\text{air}[/tex] denote the angle at which the light enters the glass block from the air.
By Snell's Law:
[tex]n_{\text{glass}} \, \sin(\theta_{\text{glass}}) = n_{\text{air}} \, \sin(\theta_{\text{air}})[/tex].
Rearrange the Snell's Law equation to obtain:
[tex]\begin{aligned} \sin(\theta_{\text{glass}}) &= \frac{n_{\text{air}} \, \sin(\theta_{\text{air}})}{n_{\text{glass}}} \\ &= \frac{(1.00)\, (\sin(45^{\circ}))}{1.50} \\ &\approx 0.471\end{aligned}[/tex].
Hence:
[tex]\begin{aligned} \theta_{\text{glass}} &= \arcsin (0.471) \approx 28^{\circ}\end{aligned}[/tex].
In other words, the angle of refraction in the glass would be approximately [tex]28^{\circ}[/tex].
Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.
15 kHz
150 kHz
1500 kHz
150 Hz
Answer: A
Explanation: We know that f=p*n
f=50*300=15000 Hz = 15kHz.
Have a great day! <3
If the number of revolutions is 300 and the paired poles are 50 , then the frequency would be 15 kHz, therefore the correct answer is option A.
What is the frequency ?It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.
The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,
F = 1 / T
As given in the problem, we have to calculate frequency if the number of revolutions is 300 and the paired poles are 50.
F = 300 × 50
= 1500 kHz
Thus, If the number of revolutions is 300 and the paired poles are 50, then the frequency would be 15 kHz, therefore the correct answer is option A.
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High-power laser in factories are used to cut through cloth and metal. One such laser has a beam diameter of 1.00mm and generates an electric field having an amplitude 0.800MV/m at the target. Find(a) the amplitude of the magnetic field produced,(b) the intensity of the laser and (c) the power delivered by the laser.
(a) The amplitude of the magnetic field produced is 2.667 mV/m.
(b) The intensity of the laser is 2.832 W/m².
(c) The power delivered by the laser is 2.22 x 10⁻⁶ W.
Amplitude of the magnetic field producedIn electromagnetic waves, the amplitude of magnetic field is the maximum field strength of the magnetic fields.
B₀ = E₀/c
where;
E₀ is the amplitude electric fieldB₀ is the amplitude magnetic fieldc is speed of lightB₀ = (0.8 MV/m) / (3 x 10⁸)
B₀ = (0.8 x 10⁶ V/m) / (3 x 10⁸)
B₀ = 2.667 x 10⁻³ V/m
B₀ = 2.667 mV/m
Intensity of laserThe intensity of the laser is calculated as follows;
I = ¹/₂ε₀E₀²
I = (0.5)(8.85 x 10⁻¹²)(0.8 x 10⁶)²
I = 2.832 W/m²
power delivered by the laserP = IA
where;
A is the area of the beamA = πd²/4
where;
d is diameterA = π(1 x 10⁻³)²/4
A = 7.854 x 10⁻⁷ m²
Power = (2.832 W/m²) x (7.854 x 10⁻⁷ m²)
Power = 2.22 x 10⁻⁶ W
Thus, the amplitude of the magnetic field produced is 2.667 mV/m.
The intensity of the laser is 2.832 W/m².
The power delivered by the laser is 2.22 x 10⁻⁶ W.
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A ship is 2.40km from a large rock cliff when it sounds its foghorn at a frequency of 178Hz. How long will it take for the sounds to return to the ship? (Sound travels 343m/sec)
A ship sounds its foghorn at a frequency of 178Hz when it is 2.40km away from a sizable rock face. (Audio travels at 343 m/s.) Reconstruction succeeded in resolving the federalism controversy that had dogged the country since the 1790s.
By most other standards, reconstruction was a failure: radical Republican legislation eventually failed to shield former slaves from white persecution and to bring about significant alterations to the South's socioeconomic structure. The federalism question, which had been a topic since the 1790s practically immediately, was at hand when President Rutherford B. Former Confederate officials and slaves returned to the South after Hayes withdrew federal soldiers from the region in 1877. These newly powerful white southern legislators established anti-progressive laws like voter ID requirements and black codes with the help of a conservative Supreme Court in an effort to roll back the rights that blacks had won during Radical Reconstruction. With its rulings in the Slaughterhouse Cases, the Civil Rights Cases, and United States v., the U.S. Supreme Court strengthened this anti-progressive movement federalism.
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The density of molybdenum is 10.28 g/cm^3 and it crystallizes in the face centered cubic unit cell. Calculate the edge length of the unit cell. (The atomic mass of Mo is 95.96 g/mole)
The edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.
Volume of molybdenumV = (zm/ρN)
where;
z is 2 for cubic unit cellm is mass of the molybdenumρ is density of the molybdenumV = (2 x 95.96) / (10.28 x 6.02 x 10²³)
V = 3.10 x 10⁻²³ cm³
Edge length of the unit cella³ = V
a = (V)^¹/₃
a = ( 3.10 x 10⁻²³)^¹/₃
a = 3.142 x 10⁻⁸ cm
a = 3.142 x 10⁻¹⁰ m
a = 314.2 x 10⁻¹² m
a = 314.2 pm
Thus, the edge length of the unit cell at the given atomic mass and density of the molybdenum is 314.2 pm.
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*According to Bohr's Theory, what is the maximum number of electron orbital layers?*
Four are known: s, p, d, and f.
Thank you,
Eddie
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
a) What is the distance of an object from the mirror's vertex if the image is real and has the same height as the object? Follow the sign rules.
The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
What is concave mirror?A concave mirror has a reflective surface that is curved inward and away from the light source.
Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.
Object distance of the concave mirrorApply mirrors formula as shown below;
1/f = 1/v + 1/u
where;
f is the focal length of the mirrorv is the object distanceu is the image distancewhen image height = object height, magnification = 1
u/v = 1
v = u
Substitute the given parameters and solve for the distance of the object from the mirror's vertex
1/f = 1/v + 1/v
1/f = 2/v
v = 2f
v = 2(19.5 cm)
v = 39 cm
Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
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A body of mass 80kg moving with a velocity of 6m/s hits a stationary body
of mass of 40kg. If the two bodies stick after the impact, calculate their
common velocity.
Answer:
4 m/s
Explanation:
Conservation of momentum , mv
Before collision 80 kg * 6 m/s = 480 kg-m/s
after the collision,
the momentum is the same but the mass is 80+40 kg =120 kg
mv = 480
120 * v = 480
v = 4 m/s
The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5 m. If the box has a mass of 1 kg, what is the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s?
The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
Acceleration of the box
The acceleration of the box is calculated as follows;
vf² = vi² + 2as
a = (vf² - vi²)/2s
a = (11.5² - 13²) / (2 x 8.5)
a = -2.16 m/s²
Time of motion of the boxThe time taken for the box to travel is calculated as follows;
a = (vf - vi)/t
t = (vf - vi) / a
t = (11.5 - 13) / (-2.16)
t = 0.69 s
Average power supplied by the frictionP = Fv
P = (ma)(vf - vi)
P = (1 x -2.16) x (11.5 - 13)
P = 3.24 W
Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.
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The gravitational force exerted by the Sun on the Earth holds the Earth in an orbit around the Sun. Let us assume that the orbit is perfectly circular. The work done by this gravitational force during a short time interval in which the Earth moves through a displacement in its orbital path is (a) zero (b) positive (c) negative (d) impossible to determine
The work done by this gravitational force during a fast time interval in which the Earth drives through a displacement in its orbital path is (a) zero.
What is the unit of gravitational force?Gravitational Force
The kilogram force (kgf) is a gravitational unit of force. The force exercised by the earth on a body of mass 1 kg is regarded as 1 kgf. Force = Mass × Acceleration. 1 kgf = 9.8 N.
The gravitation force and perpendicular to the path of movement
hence angle between gravitational force [ F ]and displacement (d)=90°
work done (W)= F d cos 90°
= 0
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In traveling to the Moon, astronauts aboard the Apollo spacecraft put spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis.
a)Determine the angular acceleration of the ship.
Express your answer using two significant figures.
b)Determine the radial component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
c)Determine the tangential component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex] , radial component is 0.016 m/[tex]s^{2}[/tex], tangential component is 0.9347*[tex]10^{-5} m/s^{2}[/tex].
Angular acceleration,
ω=1/60=0.016 rev/s
[tex]\alpha[/tex]=0.016/(6*60)=4.44*[tex]10^{-5} rev/s^{2}[/tex]
The angular acceleration is 4.44*[tex]10^{-5} rev/s^{2}[/tex]
Radial component of the linear acceleration=[tex]\alpha_{r}[/tex]
v=ωr=0.016*4.75=0.076 m/s
[tex]\alpha_{r}[/tex]=0.076 /4.75=0.016 m/[tex]s^{2}[/tex]
The tangential component = [tex]\alpha_{t}[/tex]=4.44*[tex]10^{-5} rev/s^{2}[/tex]/4.75=0.9347*[tex]10^{-5} m/s^{2}[/tex]
Angular accelerationThe temporal rate at which angular velocity changes is known as angular acceleration. Due to the fact that there are two different types of angular velocity—spin angular velocity and orbital angular velocity—there are also two different types of angular acceleration, referred to as spin angular acceleration and orbital angular acceleration, respectively. The terms "orbital angular acceleration" and "spin angular acceleration" describe the angular acceleration of a point particle about a fixed origin and, respectively, the angular acceleration of a rigid body about its center of rotation.
The unit of measurement for angular acceleration is the angle per unit time squared, or, in SI units, radians per second squared. It is typically denoted by the symbol alpha ().
In traveling to the Moon, astronauts aboard the Apollo spacecraft put spacecraft into a slow rotation to distribute the Sun's energy evenly (so one side would not become too hot). At the start of their trip, they accelerated from no rotation to 1.0 revolution every minute during a 12-min time interval. Think of the spacecraft as a cylinder with a diameter of 8.5 m rotating about its cylindrical axis.
a)Determine the angular acceleration of the ship.
Express your answer using two significant figures.
b)Determine the radial component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
c)Determine the tangential component of the linear acceleration of a point on the skin of the ship 9.5 min after it started this acceleration.
Express your answer to two significant figures and include the appropriate units.
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A -4.00 nC point charge is at the origin, and a second -6.00 nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x -axis at x = 0.200 m'
The net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
Force on electron due to charge 1
The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq₀q₁/r²
where;
k is coulomb's constantq0 is charge at the originq1 is the charge at 0.2 m (electron)r is the distance between the chargesF(01) = (9 x 10⁹ x 4 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.2²)
F(01) = 1.44 x 10⁻¹⁶ N
Force on electron due to charge 2The force exerted on the electron due to the charge q1 placed at the origin is calculated as follows;
F = kq1q2/r²
where;
k is coulomb's constantq2 is charge at the 0.8 mq1 is the charge at 0.2 m (electron)r is the distance between the charges = 0.8 m - 0.2 m = 0.6 mF(12) = (9 x 10⁹ x 6 x 10⁻⁹ x 1.6 x 10⁻¹⁹)/(0.6²)
F(12) = 2.4 x 10⁻¹⁷ N
Net force on the electronF(net) = 2.4 x 10⁻¹⁷ N + 1.44 x 10⁻¹⁶ N
F(net) = 1.68 x 10⁻¹⁶ N
Thus, the net electric force that the two charges would exert on an electron placed at point on the x -axis is 1.68 x 10⁻¹⁶ N.
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A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude of the vector? What is the angle between the vector and the positive x-axis?
[tex]\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}[/tex]
We would calculate the magnitude by applying pythagorean theorem:
[tex]\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}[/tex]
[tex]\longrightarrow \sf{Magnitude = 33.12}[/tex]
[tex]\longrightarrow \sf{The \: vector \: is \: (- 14, 30)}[/tex]
The angle between two vectors is given by the formula:
[tex]\sf{\longrightarrow \small \cos \emptyset = \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }[/tex]
In two dimensional, the x axis of vector form is:
[tex]\small\sf{\longrightarrow (b1, b2) = (1, 0) }[/tex]
[tex]\sf{\longrightarrow \small \cos \: \emptyset = \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }[/tex]
[tex]\small\longrightarrow \sf{ \dfrac{14}{33.12} }[/tex]
[tex]\small\longrightarrow \sf{\emptyset \: = arcCos (\dfrac{ - 14}{33.12} )}[/tex]
[tex]\small\longrightarrow \sf{\emptyset= 115^\circ}[/tex]
[tex]\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}[/tex]
[tex] \small\bm{The \: angle \: between \: the \: vector \: }[/tex]
[tex]\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}[/tex]
Based on the information provided, how confident can you be that this exoplanet is able to support life as we know it?
The information is missing here but data that may support the existence of life in an exoplanet may include the discovery of atmospheric biological gases.
What is an exoplanet?An exoplanet is any planet that surrounds another star beyond the sun, and therefore they are discovered by analyzing other galaxies in the Universe.
The existence of life in these exoplanets can be indirectly supported by the presence of gases that are only produced by organisms, which are detected by suitable telescopes that analyze emission spectra.
In conclusion, the information is missing here but data that may support the existence of life in an exoplanet include biological gases.
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PLEASE HELP 20 POINTS
Scientists often need to look for patterns that occur in the data they collect and analyze. Explain why identifying patterns is important, using an example from your investigation about habitable worlds.
Answer:
Patterns in science are a little different. Data doesn't have to follow a trend, always going up or down over time. A pattern is a when data repeats in a predictable way. A good example of a pattern in science comes from the father of genetics, Gregor Mendel.After data is collected, it can be analyzed by looking for trends, patterns, and relationships. Trends are general directions of data, such as an overall increase in global temperature. Patterns don't necessarily involve data going one way or the other, but rather describe a repeating observation.In order to interpret and understand scientific data, one must be able to identify the trends, patterns, and relationships in it. Examine the importance of scientific data and recognize how understanding its trends, patterns, and relationships can lead a researcher to support or refute a hypothesis. Updated: 01/06/2022 What Is Scientific Data?
Explanation:
Two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.
(A) What is the magnitude (in N) of the electric force that one particle exerts on the other?
______N
(B) Is the force attractive or repulsive?
The magnitude of the electric force that one particle exerts on the other is 8.59 x [tex]10^{-8}[/tex] N and the force is attractive.
How is Coulomb's Law Stated ?Coulomb law states that the attractive or repulsive force between two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. That is,
F = KQq / r²
Given that two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.
Where
K = constant of proportionality = 9 x [tex]10^{9}[/tex] Nm²/c²Q = 7.03 x [tex]10^{-9}[/tex] Cq = 4.02 x [tex]10^{-9}[/tex] Cr = 1.72 m(A) The magnitude of the electric force that one particle exerts on the other will be
F = (9 x [tex]10^{9}[/tex] x 7.03 x [tex]10^{-9}[/tex] x 4.02 x [tex]10^{-9}[/tex]) / 1.72²
F = 2.54 x [tex]10^{-7}[/tex] / 2.96
F = 8.59 x [tex]10^{-8}[/tex] N
(B) Since our answer is positive, the force is therefore attractive.
Therefore, the magnitude of the electric force that one particle exerts on the other is 8.59 x [tex]10^{-8}[/tex] N and the force is attractive.
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A. The magnitude (in N) of the electric force that one particle exerts on the other is 8.60×10⁻⁸ N
B. The force is repulsive
A. How to determine the magnitude of the electric forceFrom the question given above, the following data were obtained:
Charge 1 (q₁) = 7.03 nC = 7.03×10¯⁹ CCharge 2 (q₂) = 4.02 nC = 4.02×10¯⁹ CElectric constant (K) = 9×10⁹ Nm²/C²Distance apart (r) = 1.72 mForce (F) =?The magnitude of the electric force can be obtained by using the Coulomb's law equation as shown below:
F = Kq₁q₂ / r²
F = (9×10⁹ × 7.03×10¯⁹ × 4.02×10¯⁹) / (1.72)²
F = 8.60×10⁻⁸ N
B. How to determine whether the force is attractive or repulsiveFrom the question given, we were told that:
Charge 1 (q₁) = 7.03 nC Charge 2 (q₂) = 4.02 nCSince both charge are positive, then the force attraction between them is repulsive as like charges repels and unlike charges attracts
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A sine bar is used to determine the angle of a part feature. The length of the sine
bar=8.000in. The rolls have a diameter of 1.000 in. All inspection is performed on a surface plate. In
order for the sine bar to match the angle of the part, the following gage blocks must be stacked: 2.0000,
0.5000, 0.2500, and 0.0050. Determine the angle of the part feature.
A sine bar is used to get the angular measurement of a part feature. The angle of the part feature is 18.24°
From the question
Given that,
Length of the sine bar, L = 8.000 in
Diameter of the rolls = 1.000 in
Height under the roll, H = 2.0000 + 0.5000 + 0.0050
= 2.505 in
From sine bar formula,
we know that,
H = sin A x L
sin A = H ÷ L
where,
A ⇒ angle of part feature
H ⇒ height under the roll
L ⇒ length of the sine bar
Substituting values in the above equation,
sin A = H / L
A = sin⁻¹ ( 2.505 ÷ 8 )
A = sin⁻¹ (0.3131)
A = 18.24⁰
Hence the angle of the part feature = 18.24°
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If A = 6î - 8ĵ, B = −8î + 3ĵandC = 26î + 19ĵ. Find a and b Such that aẢ +bB + C = 0
Sol^n :
aA=6ai-8aj
bB=-8bi-3bj
C=26i-19j
Given aA+bB+C=0i+0j
(6a-8b+26)i-(8a+3b+19)j=0i+0j
comparing coefficient of i and j
6a-8b=-26
2a-4b=-13.....(1)
8a+3b=-19......(2)
making the coefficient of a similar
8a-16b=-52....(3)
8a+3b=-19....(4)
equation 4-3 implies
19b=33
b=33/19
from (1)
2a=-13+4(33/19)
2a=-115/19
a=-115/38
Give two examples of situations or application where electronic circuits are used
2. What energy source produces most of the electrical energy in the United States?
natural gas
O oil
O coal
Answer:
Nat gas
Explanation:
Approx:
Natural gas 38% coal 21% oil 19%
A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?
The values of the required solutions are
F= 53731 N N=37.05What is the minimum number of horses required?What is Force?
In the field of physics, an influence that can alter the motion of an object is referred to as a force. An object having mass can experience a change in its velocity, often known as an acceleration, when subjected to a force. Intuitively, force can also be conceptualized as either a push or a pull. Because it may be measured in both magnitude and direction, a force is considered a vector quantity.
What is atmospheric pressure?
The pressure that is exerted within the atmosphere of the Earth is referred to as barometric pressure as well as atmospheric pressure. As a unit of pressure, the standard atmosphere is defined as having a value of 101,325 Pa. This value is equivalent to 1013.25 millibars, 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.
In most situations, the equation for force can be expressed numerically as
Force = pressure * A
Therefore
[tex]F= (970 - 15 B)(\pi * (0.430 )^2)[/tex]
F= 53731 N
In conclusion, If each horse can pull with a force of 1450N
The number of horses required is
[tex]N=\frac{60754 }{1450}[/tex]
N=37.05
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A child rolls a ball on a level floor 5.2 m to another child.
If the ball makes 15.0 revolutions, what is its diameter?
Express your answer to two significant figures and include the appropriate units.
Diameter of the ball = 11.03cm
The ball rolled by the child must cover a distance in a linear motion which will be then equal to the circumference of the ball after one complete revolution.
Linear distance = 5.2m
d = diameter of the ball =?
Total revolutions = 15
S = 2πr
S = πd
For 15 revolutions
S = 15πd
The Diameter of a ball is double the radius of the ball thus we use D instead of 2r in the formula
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A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .
If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Express your answer to two significant figures and include the appropriate units.
The body moves at a velocity of 1.62m/s after the bullet emerges.
Given:Mass of bullet, [tex]m_1[/tex] = 22g
= 0.022 kg
Mass of the block, [tex]m_2[/tex] = 1.9 kg
Velocity of bullet , [tex]v_1[/tex] = 265 m/s
[tex]v_2 = 0[/tex]
According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.
After penetration;
[tex]v^{'}_1 =125 m/s[/tex]
[tex]v^{'}_2=?[/tex]
The formula for calculating the collision of a body is expressed as:
p = mv
m is the mass of the body
v is the velocity of the body
∴ Momentum before = Momentum after
Substitute the given parameters into the formula as shown:
[tex]m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s[/tex]
Therefore, It moves with a velocity of 1.62 m/s.
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HELP PLS
The minimum frequency for the photoelectric effect of a zinc plate is 9.7 × 10^14 Hz. If a green photon with a wavelength of 532nm strikes the plate, will electrons be emitted? Show your work to support your answer.
yes electrons will be emitted with the kinetic energy of
[tex]30.87 \times 10 { }^{ - 19} j[/tex]
what is photoelectric effect?
In photoelectric effect, the electrically charged particles are released from or within a material when it absorbs electromagnetic radiation.
what is kinetic energy?
It is a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy.
Given:
wavelength,λ
[tex] = 5.32 \times 10 {}^{ - 8} m[/tex]
Frequency
[tex] = 9.7 \times {10}^{14} [/tex]
Einstein's photoelectric equation
€= w + KE
€ = hc/ λ
[tex] = \frac{6.63 \times {10}^{ - 34} \times 3 \times {10}^{8} } {5.32 \times 10 {}^{ - 8} } [/tex]
[tex]€= 3.73 \times {10}^{ - 18} [/tex]
w=h.f
[tex] = 6.63 \times {10}^{ - 34} \times 9.7 \times {10}^{14} [/tex]
[tex] = 6.43 {10}^{ - 19} j[/tex]
KE = € - W
[tex]37.3 \times {10}^{ - 19} - 6.43 \times {10}^{ - 19} [/tex]
[tex]kE = 30.87 \times {10}^{ - 19} [/tex]
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the apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ? use negative is the acceleration vector is pointing downwards
Answer:
-.457 m/s^2
Explanation:
Actual weight = 60 .3 (9.81) = 591.54 N
Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift
60.3 ( 9.81 - L ) = 564
solve for L = .457 m/s^2 DOWNWARD
so L = - .457 m/s^2
If a person walks first 70 m in the direction 37° north of east, and then walks 82 m in the
direction 20° south of east, and finally walks 28 m in the direction 30° west of north.(2pt)
a) How far and at what angle is the Aster's final position from her initial position?
b) In what direction would she has to head to return to her initial position
The Aster's final position from her initial position is determined as 45.3 m.
The direction that she has to head to return to her initial position is 30° west of north.
Distance of the person from her initial position
The distance traveled by the person can be determined by making a sketch of the person's journey.
From the initial distance traveled and the second distance distance the angle between the two position is calculated as;
θ = 37⁰ + 20⁰ = 57⁰
Distance opposite to the angleThe distance opposite to the angle is resultant displacement and it is calculated as follows;
r² = (70²) + (82²) - (2 x 70 x 82) x cos(57)
r² = 5,371.54
r = √5,371.54
r = 73.3 m
Haven walked 28 m in the same direction to her initial position, the remaining distance is calculated as follows;
d = 73. 3 m - 28 m
d = 45.3 m
Thus, the Aster's final position from her initial position is determined as 45.3 m.
The direction that she has to head to return to her initial position is 30° west of north.
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Answer:
See below
Explanation:
I will break the legs into vertical and horizontal components then add them
Horizontal = 70 cos 37 + 82 cos 340 + 28 cos 120 = 118.96 m
Vertical = 70 sin 37 + 82 sin 340 + 28 sin 120 = 38.33 m
Resultant distance from origin is found via pythag theorem
d^2 = 118.96^2 + 38.33^2 d = 124.98 m
direction FROM origin is arctan ( 38.33/118.96) = 17.86°
to get BACK to the origin , will have to walk 180 degrees from this = 197.86°
The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
a.
i. PU is greater than PQ.ii. VU is Greater than 2VTb.
i. PR is Equal to PU.ii. VR is Equal to VSc.
i. VQ is Equal to VUii. PR is Greater than PS.What is pressure?Pressure is the force per unit area on a surface.
What is speed?Speed is the distance moved per unit time.
Pressure
Since pressure, P = hρg where
h = depth, ρ = density of liquid and g = acceleration due to gravity.Since ρ and g are constant
P ∝ h
So, we see that pressure is directly proportional to depth.
a. i. Pressure between R and USince U is lower than Q, Pressure at U is greater than pressure at Q.
So,PU is greater than PQ.
ii. Speed between U and TUsing the continuity equation
VUAU = VTAT where
VU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmVUT= speed at T, AT = cross-sectional area at T = π(dT)² where dT = diameter at T = 2.0 mmSo, VUAU = VTAT
VUπ(dU)² = VTπ(dT)²
VU = VT(dT)²/(dU)²
VU = VT(2.0)²/(1.0)²
VU = VT(4)
VU = 4VT
Since VU = 4VT,VU is Greater than 2VT
b i. Pressure between R and USince R is at the same depth as U, Pressure at R is equal to pressure at U.
So,PR is Equal to PU.
ii. Speed between R and SUsing the continuity equation
VRAR = VSAS where
VR = speed at R, AR = cross-sectional area at R = π(dR)² where dR = diameter at R = 2.0 mmVS= speed at S, AS = cross-sectional area at S = π(dS)² where dS = diameter at S = 2.0 mmSo, VRAR = VSAS
VRπ(dR)² = VSπ(dS)²
VR = VS(dS)²/(dS)²
VR = VS(2.0)²/(2.0)²
VR = VS(1)
VR = VS
Since VR = VS,VR is Equal to VS
c. i. Speed between Q and UUsing the continuity equation
VQAQ = VUAU where
VQ = speed at Q, AQ = cross-sectional area at Q = π(dQ)² where dQ = diameter at Q = 1.0 mmVU = speed at U, AU = cross-sectional area at U = π(dU)² where dU = diameter at U = 1.0 mmSo, VQAQ = VUAU
VQπ(dQ)² = VUπ(dU)²
VQ = VU(dU)²/(dQ)²
VQ = VU(1.0)²/(1.0)²
VQ = VU(1)
VQ = VU
Since VQ = VU, VQ is Equal to VU
Ii. Pressure between R and SSince R is lower than S, Pressure at R is greater than pressure at S.
So,PR is Greater than PS.
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A thin-walled hollow sphere has a radius 4cm from the center
of the sphere, the eletric field points radially inward and
has a magnitude of 1.5 × 10^4 NC". How much charge is on the surface
The electric charge on surface is 2.66×10-⁹NC
What is an electric charge?The electric charge is the property of subatomic particle that causes it to experience a force when placed in an electric and magnetic field.
We have given ,
Electric field= 1.5×10⁴N/C
Radius = 4cm = 0.04m
E = k× q/ r²
1.5×10⁴×16×10-⁴/9×10⁹=q
q= 2.66×10-⁹C
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Lam Lesson Name: Uncovering Your Personality
m number: 700047RR
Exam Guidelines
Exam Instructions
Question 10 of 20:
Select the best answer for the question.
10. Which characteristic of a turbulent person causes them to always strive for self-improvement, and to never see accomplishing a goal as good enough?
O A. Confident
B. Self-critical
O C. Ignorant
O D. Calm
Mark for review (Will be highlighted on the review page)
ex Previous Question
Next Questin
Review My F
The correct answer is Self-critical.
Why is self-improvement?
Enhancing strengths, mental health, and even mending relationships benefit self-improvement. Simple actions like reading a book, trying something new, meditating, or even getting up early are some ways to improve oneself. There are so many easy, efficient methods to begin the process of improving oneself.A self-improvement strategy enables you to build the life you want for yourself. It enables you to maintain perspective on your priorities and the things most important to you in life to experience greater meaning and fulfillment.Self-development is taking steps to better yourself, such as by learning new skills or overcoming bad habits. An example of self-development is taking courses at the university to learn new skills and interesting things.Self-critical:
Self-critical causes them to always strive for self-improvement and never to see accomplishing a goal as good enough.
The characteristic of a turbulent person causes them to always strive for self-improvement and to never see accomplishing a goal as good enough is Self-critical.
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The figure illustrates flow through a pipe with diameters of 1.0 mm and 2.0 mm and with different elevations. Px is the pressure in the pipe, and Vx is the speed of a non-viscous incompressible fluid at locations x = Q,R,S,T, or U. Options are: Greater than, Less than, Equal to
PU is ... PQ
VU is ... 2VT
PR is ... PU.
VR is ... VS
VQ is ... VU
PR is ... PS
By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,
[tex]P_u > P_q[/tex] [tex]V_u > 2V_t[/tex] [tex]P_u > P_r[/tex] [tex]V_r=V_s[/tex] [tex]V_q=V_u[/tex][tex]P_r > P_s[/tex]To find the answer, we need to know about the Bernoulli's equation as well as the Equation of continuity.
How to find the solution?1) We have to find the relation between pressure at U and Q.
We have the Bernoulli's equation,[tex]P+\frac{1}{2}dV^2+dgh=constant.[/tex]
where, P is the pressure, V is the velocity, d is density, g is acceleration due to gravity and h is the height of the flow.
By using the equation, we can find the pressure at U and Q.[tex]P_u+\frac{1}{2}dV_u^2+dg*0= P_q+\frac{1}{2}dV_q^2+dgh\\where,\\V_u=V_q, \\since, D_p=D_q\\Thus,\\P_u=P_q+dgh[/tex]
[tex]P_u > P_q[/tex]
2) We have to find the relation between velocity at U and T.
For this, we have the equation of continuity as,[tex]AV=constant\\A_1V_1=A_2V_2[/tex]
From the diagram, we have,[tex]A_u=\pi r^2=\pi *(0.5)^2=0.25\pi *mm^2\\A_t=\pi *1=\pi mm^2\\V_u=V\\V_t=?[/tex]
Thus, the relation between velocity at U and T is,[tex]V_t=\frac{A_uV_u}{A_t}=\frac{V}{4}[/tex]
[tex]2V_t=2*\frac{V}{4} =\frac{V}{2}\\V_u=V\\[/tex]
[tex]V_u > 2V_t[/tex]
3) We have to find the relation between pressure at R and U
[tex]P_u+\frac{1}{2}dV_u^2+dg*0= P_r+\frac{1}{2}dV_r^2+dg*0\\\\V_u=V , then\\V_r=V_t=\frac{V}{4}=\frac{V_u}{4} \\\\P_u=P_r+\frac{1}{16}[/tex]
[tex]P_u > P_r[/tex]
4) We have to find the relation between velocity at R and S
Both points R and S, have same area, thus same velocity.[tex]V_r=V_s[/tex]
5) We have to find the relation between velocity at Q and U
Both points Q and U, have same area, thus same velocity.[tex]V_q=V_u[/tex]
6) We have to find the relation between pressure at R and S
Both points R and S, have same area and thus, same velocities.[tex]P_r=P_s+dgh[/tex]
[tex]P_r > P_s[/tex]
Thus, we can conclude that, By Bernoulli's equation as well as the Equation of continuity the relation between pressure and velocities at different points are,
[tex]P_u > P_q[/tex] [tex]V_u > 2V_t[/tex] [tex]P_u > P_r[/tex] [tex]V_r=V_s[/tex] [tex]V_q=V_u[/tex][tex]P_r > P_s[/tex]Learn more about the Bernoulli's equation here:
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