Complete Question
Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.
Answer:
[tex]I=0.032kgm^2[/tex]
Explanation:
From the question we are told that:
Mass[tex]m=0.2kg[/tex]
Length [tex]l=0.4m[/tex]
Generally the equation for Inertia is mathematically given by
[tex]I=md^2[/tex]
[tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]
[tex]I=0.032kgm^2[/tex]
Polarized sunglasses:
a. block most sunlight because sunlight is polarized
b. are better but work the same way as non-polarized sunglasses
c. are polarized to filter out certain wavelengths of light
d. block reflected light because reflected light is partially polarized.
Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.
The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.
Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.
Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.
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Kelsey the triathelete swims 1.5 km east, then bikes 40 km north, and then runs 10 km west. Which choice gives the
correct solution for the resultant?
R2 = 402 – 8,52
R2 = 402 - 102 - 2(40)(1.5) cos 10
R2 = 102 - 40
R2 = 10- - 402 – 2(1.5)(10) cos 40
Answer:
Hey,. its a simple question. hope you learn from the solution. check attached picture
Explanation:
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone subsequently falls to the ground, which is 14.5 m below the point where the stone leaves his hand.
At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity.
How much time is the stone in the air?
elapsed time:
Answer:
Speed=28.1m/s(to 3s.f.) , Time=2.19s(to 3s.f.)
Explanation:
Time=Distance/Speed
=14.5/6.63
=2.19s(to 3s.f.)
Acceleration=Final Velocity(v)-Initial Velocity(u)/Time
9.81=v-6.63/2.19
v-6.63=21.5
v=28.1m/s
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common isotope, U-238. Estimate the radius of the circle traced out by a singly ionized lead atom moving at the same speed.
Answer:
21.55 m
Explanation:
Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.
a. This low value is possible because the magnetic field of Saturn is so strong.
b. This low value is possible because the magnetic field of Saturn is so weak.
c. This low value is possible because the density of Saturn is so high.
d. This low value is possible because the density of Saturn is so low.
Answer:
Explanation:
That is an amazing fact.
The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.
The answer is D
If the mass of an object is 10 kg and the
velocity is -4 m/s, what is the momentum?
A. 4 kgm/s
B. -40 kgm/s
C.-4 kgm/s
D. 40 kgm/s
Answer:
B. -40 kgm/s is the answer
answer bhejo please please please
Answer:
Various uses of water :
1. Water is used for daily purpose like cooking , bathing , cleaning and drinking.
2. Water used as a universal solvent.
3. water maintains the temperature of our body.
4. Water helps in digestion in our body.
5 .water is used in factories and industries.
6. Water is used to grow plants , vegetables and crops.
State TRUE or FALSE.
1. We use muscular force to lift a bucket of water.
2. A bow uses mechanical force of the bow string to shoot an arrow.
3. The force of friction enables us to walk on earth.
4. Plants use solar energy to make their food.
5. The energy stored inside the earth is called atomic energy
Answer:
1. True
2. False
3. True
4. True
5. True
Answer:
that is pure falsereeeeeeeee
Explanation:
If an object with constant mass is accelerating, what does Newton's second
law imply?
A. It will continue to accelerate until it meets an opposing force.
B. The object is exerting an opposite but equal force.
C. A force must be acting on the object.
D. The object will be difficult to decelerate.
Answer:
C. A force must be acting on the object.
Explanation:
This is due to the action of its momentum direction.
[tex].[/tex]
Define wave length as applied to wave motion
Answer: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.
Explanation:
Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.
How can I solve the following statement?
What is the magnitude of the electric field at a point midway between a −8.3μC and a +7.8μC charge 9.2cm apart? Assume no other charges are nearby.
Answer:
The net electric field at the midpoint is 6.85 x 10^7 N/C.
Explanation:
q = − 8.3 μC
q' = + 7.8 μC
d = 9.2 cm
d/2 = 4.6 cm
The electric field due to the charge q at midpoint is
[tex]E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C[/tex] leftwards
The electric field due to the charge q' at midpoint is
[tex]E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C[/tex]
The resultant electric field at mid point is
E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
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Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pulls on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Answer:
The answer is "Option b, c, and a".
Explanation:
Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.
We now understand the effects of gravity:
[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]
The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:
b. Pull-on phone from earth
c. Pull-on phone from sun
a. Pull phone from you
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound
The question is incomplete. The complete question is :
Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.
What is the frequency of the sound?
Solution :
Given :
The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]
The speaker are in phase and so the path difference is zero constructive interference occurs.
At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]
Therefore,
[tex]$AD-BD = \frac{\lambda}{2}[/tex]
[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]
[tex]$\lambda = 2 \times 0.4985$[/tex]
[tex]$\lambda = 0.99714 \ m$[/tex]
Thus the frequency is :
[tex]$f=\frac{v}{\lambda}$[/tex]
[tex]$f=\frac{340}{0.99714}$[/tex]
[tex]f=340.9744[/tex] Hz
two object A and B vertically thrown upward with velocities of 80m/s and 100m/s at two seconds interview where and when will the two object meet.
Answer:
THIS IS YOUR ANSWER:
☺✍️HOPE IT HELPS YOU ✍️☺
What is the magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away?
Answer:
Force,
[tex]F = \frac{kQ_{1} Q_{2} }{ {r}^{2} } \\ F = \frac{(9 \times {10}^{9}) \times (25 \times {10}^{ - 6}) \times (10 \times {10}^{ - 6} ) }{ {(0.85)}^{2} } \\ \\ F = 3.114 \: newtons[/tex]
The magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away would be 311.4 N.
What is Coulomb's Law?Coulomb's law can be stated as the product of the charges and the square of the distance between them determine the force of attraction or repulsion acting in a straight line between two electric charges.
The math mathematical expression for the coulomb's law given as
F= k Q₁Q₂/r²
where F is the force between two charges
k is the electrostatic constant which is also known as the coulomb constant,it has a value of 9×10⁹
Q₁ and Q₂ are the electric charges
r is the distance between the charges
As given in the problem two charges a 25μC charge exerts on a -10μC charge 8.5cm away
By substituting the respective values in the above formula of Coulomb law
F =9×10⁹×(25×10⁻⁶)×(-10×10⁻⁶)/(8.5×10⁻²)²
F= -311.4 N
A negative sign represents that the force is attractive in nature
Thus, the magnitude of the force is 311.4 N.
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Joule is a SI unit of power
Measuring cylinder is used to measure the volume of a liquid
Answer:
The SI unit of power is watt
A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 ksi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 ksi. Find the factor of safety guarding against a static failure, and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis use Modified Goodman criterion.
Answer:
The correct solution is:
(a) 1.66
(b) 1.05
Explanation:
Given:
Bending stress,
[tex]\sigma_b = 25 \ kpsi[/tex]
Torsional stress,
[tex]\tau= 15 \ kpsi[/tex]
Yield stress of steel bar,
[tex]\delta_y = 60 \ kpsi[/tex]
As we know,
⇒ [tex]\sigma_{max}^' \ = \sqrt{\sigma_b^2 + 3 \gamma^2}[/tex]
[tex]= \sqrt{(25)^2+3(15)^2}[/tex]
[tex]=36.055 \ kpsi[/tex]
(a)
The factor of safety against static failure will be:
⇒ [tex]\eta_y = \frac{\delta_y}{\sigma_{max}^'}[/tex]
By putting the values, we get
[tex]=\frac{60}{36.055}[/tex]
[tex]=1.66[/tex]
(b)
According to the Goodman line failure,
[tex]\sigma_a = \sigma_b = 25 \ kpsi[/tex]
[tex]S_e = 40 \ kpsi[/tex]
[tex]\sigma_m = \sqrt{3} \tau[/tex]
[tex]=\sqrt{3}\times 15[/tex]
[tex]=26 \ kpsi[/tex]
[tex]Sut = 80 \ kpsi[/tex]
⇒ [tex]\frac{\sigma_a}{S_e} +\frac{\sigma_m}{Sut} =\frac{1}{\eta_y}[/tex]
[tex]\frac{25}{40}+\frac{26}{80}=\frac{1}{\eta_y}[/tex]
[tex]\eta_y = 1.05[/tex]
II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49. ( ) If the buckets are at rest, what is the tension in each cord? ( ) If the two buckets are pulled upward with an acceleration of 1.25 m/s by the upper cord, calculate the tension in each cord
Answer:
Here , mass of bucket ,m = 3.2 Kg
Now , let the tension in upper rope is T1
the tension in the middle rope is T2
a)
For lower bucket, balancing forces in vertical direction
T2 - mg = 0
T2 = mg
T2 = 3.2 *9.8
T2 = 31.36 N
tension in the middle rope is 31.36 N
For the upper bucket , balancing forces in vertical direction
T1 - T2 - mg = 0
T1 = T2 + 3.2 *9.8
T1 = 62.72 N
the tension in the upper rope is 62.72 N
B)
for a = 1.25 m/s^2
Using second law of motion ,for both the buckets
Fnet = ma
T1 - 2mg = 2m*a
T1 = 2*3.2*(9.8 +1.25)
T1 = 70.72 N
the tension in the upper rope is 70.7 N
Now , the lower bucket
Using second law of motion,
T2 - mg = ma
T2 = 3.2 * (9.8 + 1.25)
T2 = 35.36 N
the tension in the lower rope is 35.36 N
You are trying to push a 14 kg canoe across a beach to get it to a lake. Initially, the canoe is at rest, and you exert a force over a distance of 5 m until it has a speed of 2.1 m/s.
a. How much work was done on the canoe?
b. The coefficient of kinetic friction between the canoe and the beach is 0.2. How much work was done by friction on the canoe?
c. How much work did you perform on the canoe?
d. What force did you apply to the canoe?
Answer:
I hope this is correct!
Explanation:
a) work = (force)(displacement)
We know that d = 5m, now we just need to find the force
We need to calculate the acceleration first using a = v^2 - u^2 / 2d
final velocity (v) = 2.1 m/s , initial velocity (u) = 0 m/s , displacement (d) = 5m
a = 2.1^2 - 0^2 / 2(5)
= 0.441 m/s^2
Now we can find force using f = ma
f = (14kg)(0.441m/s^2)
= 6.174 newtons
Finally, you can calculate work now that you have force!
work = (6.174n)(5m)
= 30.87 J (you may need to round sig figs!)
b) Work done by friction = (coefficient of kinetic friction)(mg)(v)
m = 14kg, g = 9.8, coefficient of friction = 0.2
force due to friction = (0.2)(14kg x 9.8)(2.1)
= 57.624 J (again you may need to round to sig figs)
c) Work you perform = total work in direction of motion + work done by friction
total work = 30.87 J, work done by friction = 57.624 J
Work you perform = 30.84 + 57.624
= 88.464 J
d) Force you applied = work you performed / distance
work you performed = 88.464 / 5 m
= 17.6929 N (again you may need to round to sig figs)
Hope this helps ya! Best of luck <3
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image. (3)
Answer:
I think 9.5
Explanation:
............
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
a) The mass flow rate through the nozzle can be calculated with the following equation:
[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]
Where:
[tex]v_{i}[/tex]: is the initial velocity = 20 m/s
[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²
[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³
[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]
[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]
[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]
Therefore, the exit area of the nozzle is 23.6 cm².
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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.
We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:
[tex]\dot m_{in} = \dot m_{out}[/tex] (1)
Where:
[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.
[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.
Given that air flows at constant rate, we expand (1) by dimensional analysis:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)
Where:
[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.
[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.
[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.
a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:
[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]
[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]
[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]
The mass flow rate through the nozzle is 0.265 kilograms per second.
b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]
[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]
[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]
[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]
[tex]A_{out} = 23.202\,cm^{2}[/tex]
The exit area of the nozzle is 23.202 square centimeters.
if C is the vector sum of A and B C=A+B what must be true about directions and magnitude of A and B if C=A+B? what must be true about the directions and magnitude of A and B if C=0
The vector sum is the algebraic sum if the two vectors have the same direction.
The sum vector is zero if the two vectors have the same magnitude and opposite direction
Vector addition is a process that can be performed graphically using the parallelogram method, see attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.
There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel
case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector
case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.
In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector
a) Magnitudes: [tex]\| \vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| \ge 0[/tex]; Directions: [tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex], [tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex], [tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].
b) Magnitudes: [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| = 0[/tex]; Directions: [tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex], [tex]\theta_{C}[/tex] is undefined.
a) Let suppose that [tex]\vec A \ne \vec O[/tex], [tex]\vec B \ne \vec O[/tex] and [tex]\vec C \ne \vec O[/tex], where [tex]\vec O[/tex] is known as Vector Zero. By definitions of Dot Product and Inverse Trigonometric Functions we derive expression for the magnitude and directions of [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex]:
Magnitude ([tex]\vec A[/tex])
[tex]\|\vec A\| = \sqrt{\vec A\,\bullet\,\vec A}[/tex]
[tex]\| \vec A\| \ge 0[/tex]
Magnitude ([tex]\vec B[/tex])
[tex]\|\vec B\| = \sqrt{\vec B\,\bullet\,\vec B}[/tex]
[tex]\|\vec B\| \ge 0[/tex]
Magnitude ([tex]\vec C[/tex])
[tex]\|\vec C\| = \sqrt{\vec C\,\bullet \,\vec C}[/tex]
[tex]\|\vec C\| \ge 0[/tex]
Direction ([tex]\vec A[/tex])
[tex]\vec A \,\bullet \,\vec u = \|\vec A\|\cdot \|u\|\cdot \cos \theta_{A}[/tex]
[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|\cdot \|u\|}[/tex]
[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|}[/tex]
[tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex].
Direction ([tex]\vec B[/tex])
[tex]\vec B\,\bullet \, \vec u = \|\vec B\|\cdot \|\vec u\| \cdot \cos \theta_{B}[/tex]
[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex]
[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|}[/tex]
[tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex].
Direction ([tex]\vec C[/tex])
[tex]\vec C \,\bullet\,\vec u = \|\vec C\|\cdot\|\vec u\|\cdot \cos \theta_{C}[/tex]
[tex]\theta_{C} = \cos^{-1}\frac{\vec C\,\bullet\,\vec u}{\|\vec C\|\cdot\|\vec u\|}[/tex]
[tex]\theta_{C} = \cos^{-1} \frac{\vec C\,\bullet\,\vec u}{\|\vec C\|}[/tex]
[tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].
Please notice that [tex]\vec u[/tex] is the Vector Unit.
b) Let suppose that [tex]\vec A \ne \vec O[/tex] and [tex]\vec B \ne \vec O[/tex] and [tex]\vec C = \vec O[/tex]. Hence, [tex]\vec A = -\vec B[/tex]. In other words, we find that both vectors are antiparallel to each other, that is, that angle between [tex]\vec A[/tex] and [tex]\vec B[/tex] is 180°. From a) we understand that [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], but [tex]\|\vec C\| = 0[/tex].
Then, we have the following conclusions:
Magnitude ([tex]\vec A[/tex])
[tex]\|\vec A\| \ge 0[/tex]
Magnitude ([tex]\vec B[/tex])
[tex]\|\vec B\| \ge 0[/tex]
Magnitude ([tex]\vec C[/tex])
[tex]\|\vec C\| = 0[/tex]
Directions ([tex]\vec A[/tex], [tex]\vec B[/tex]):
[tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex]
Direction ([tex]\vec C[/tex]):
Undefined
Describe how the words Science and optics would appear when viewed in a plane mirror?
Answer:
Lateral inversion will occur in a plane mirror.
Explanation:
When words are displayed in a plane or flat mirror, the result is that if the words are displayed left, they change to right and if they were normally displayed right, they change to left. This phenomenon is known as lateral inversion. So, this will apply to the words, Science and optics. Only the sides will be interchanged.
A plane mirror reflects light, therefore, the image that is produced by it remains the same size. The image produced will not appear upside down. Only the sides will be interchanged.
In a photoelectric effect experiment, it is observed that violet light does not eject electrons from a particular metal. Next, red light with the same intensity is incident on the same metal. Which result is possible
Answer:
No ejection of photo electron takes place.
Explanation:
When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.
The minimum energy required to just eject an electron is called work function.
The photo electric equation is
E = W + KE
where, E is the incident energy, W is the work function and KE is the kinetic energy.
W = h f
where. h is the Plank's constant and f is the threshold frequency.
Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum uncertainty in the electron's momentum
Answer:
Uncertainty in position (∆x) = 24 × 10⁻¹⁵ mUncertainty in momentum (∆P) = ?Planck's constant (h) = 6.26 × 10⁻³⁴ Js[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
An object moving with initial velocity 10 m/s is subjected to a uniform acceleration of 8 m/s ^² . The displacement in the next 2 s is: (a) 0m (b) 36 m (c) 16 m (d) 4 m
