Answer:
6.62m/s
Explanation:
Given data
M1= 120kg
U1= 5.5m/s
M2= 105kg
U2= 7.9m/s
The system experiences an inelastic collision, the expression for inelastic collision is
M1U1+ M2U1= (M1+M2)V
Subsitute
120*5.5+ 105*7.9= (120+105)*V
660+ 829.5= (225)*V
1489.5= (225)*V
Divide both sides by 225
V= 1489.5/225
V= 6.62m/s
Hence the common velocity is 6.62m/s
1. What exactly is science?
Write the definition of
science in your own words.
Answer:
A practical activity about the study of the structure of the natural and physical world that can be shown thorough experiments and study's.
Explanation:
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension
The question is incomplete. The complete question is :
Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.
Solution :
Given :
[tex]$E_{glass fibre}$[/tex] = 200 GPa
[tex]$V_{glass fibre} = 82\%$[/tex]
[tex]$E_{epoxy}$[/tex] = 38 GPa
[tex]$V_{epoxy} = 82\%$[/tex]
Edge length = 56 mm
Cube is loaded in axial tension such that the force is uniformly applied over a cube face.
[tex]$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$[/tex]
[tex]$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$[/tex]
[tex]$=113.16 $[/tex] GPa
Applied stress [tex]$=\frac{\text{applied load}}{\text{area}}$[/tex]
[tex]$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$[/tex]
= 32.5 MPa
By Hooke's law
[tex]$\sigma = E . \epsilon$[/tex]
[tex]$\sigma = E. \frac{\Delta l}{l}$[/tex]
[tex]$\Delta l = \frac{\sigma}{E}\times l$[/tex]
Length change, [tex]$\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$[/tex]
[tex]$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$[/tex]
= 0.016 mm
Please help me ASAP!!
Answer:
the ball lose kentic energy and gains potential energy rolling upward
Answer:
c
Explanation:
when a ball is rolling down it loses potential and gains kinetic, but in this case since the ball is going upwards it is losing kinetic energy and gaining potentail like a roller coaser
the path of the earth around the sun is called
Answer:
Orbit.Explanation:
An orbit is a regular, repeating path that one object in space takes around another one, it's an imaginary path of the Earth around the Sun.
a space probe with a mass of 4000 kg expels 3,500 of its mass at a velocity of 2000 m/s. what is the velocity of the remaining 500 kg of the probe
Answer:
4.16×103 m/s
Explanation:
What is the centripetal force for a roller coaster if the mass is 10 kg and the normal force is 25 N?
Answer:
Fc = 123 Newton
Explanation:
Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.
Mathematically, net force is given by the formula;
[tex] Fnet = Fapp + Fg[/tex]
Where;
Fnet is the net force.
Fapp is the applied force.
Fg is the force due to gravitation.
Given the following data;
Normal force = 25N
Mass = 10kg
To find the centripetal force;
From the net force, we have the following formula;
Fc = N + mg
Where;
Fc is the centripetal force.
N is the normal force.
mg is the the weight of the object.
Substituting into the formula, we have;
Fc = 25 + 10(9.8)
Fc = 25 + 98
Fc = 123 Newton
[tex] \sf{ What \: is \: a \: parallel \: circuit? }[/tex]
[tex] \\ \\ \\ \\ \\ [/tex]
Answer:
parallel circuit is a kind of circuits in which all components are connected between the same two sets of electrically common points, creating multiple paths for the current to flow from one end of the battery to the other:
Answer:
Gadhe hote hai log and usme sabse pehele aati hai Jonain jaise log waah waah xD
Brainliest krdena warna chappal padenge -,-
what type of signal is utilized by the GP's satellite ?
Answer:
Hi how are you doing today Jasmine
Answer:
The answer is B
Explanation:
I did the USA test prep
NO LINKS: Which of the following best demonstrates Newton's Third Law?
Gravity is a force pulling down on you.
The ahrder you hit a volleyball the faster it will get over the net
Air resistance slows down a parachutist
Your foot pushing on the floor and the floor pushing back
The best demonstration that applies to Newton's Third Law of motion would be D) When you walk your foot pushes down on the ground while the ground pushes back on your foot.
------------------
Newton's Third Law states that for every action, there is an equal and opposite reaction. This is actually explains that forces come in pairs and forces are an interaction between two objects. As per the correct option given in the question explains Newton’s Third Law.
------------------
When you walk your foot (say object A) pushes down on the ground while the ground (say object Q) pushes back on your foot with the same force but in the opposite direction.
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below
This question is incomplete, the complete question is;
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) What is the velocity of the top of the ladder when the base is given below
7 feet away from the wall
15 feet away from the wall
20 feet away from the wall
Answer:
- When x is 7 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -7/12 ft/s
- When x is 15 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -3/2 ft/s
- When x is 20 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -8/3 ft/s
Explanation:
Given the data in the question;
Let [tex]r[/tex]ft and [tex]x[/tex]ft represent the distance of the top of ladder from the ground and the distance of the bottom of ladder from the wall respectively.
Now using the Pythagoras Theorem ( right angled triangle ).
(hypotenuse)² = ( adjacent )² + ( opposite )²
so
r² = ( 25 )² - x²
r² = 625 - x² ------- let this be equation 1
now, we differentiate with respect to t
2r [tex]\frac{dr}{dt}[/tex] = 0 - 2x [tex]\frac{dx}{dt}[/tex]
r [tex]\frac{dr}{dt}[/tex] = -x [tex]\frac{dx}{dt}[/tex]
[tex]\frac{dr}{dt}[/tex] = -x [tex]\frac{dx}{dt}[/tex] × 1/r
[tex]\frac{dr}{dt}[/tex] = -x/r [tex]\frac{dx}{dt}[/tex]
Now, from equation, r² = 625 - x²
r = √( 625 - x² )
Then
[tex]\frac{dr}{dt}[/tex] = -x / √( 625 - x² ) [tex]\frac{dx}{dt}[/tex]
given that, The base of the ladder is pulled away from the wall at a rate of 2 feet per second i.e [tex]\frac{dx}{dt}[/tex] = 2 ft/s
so;
[tex]\frac{dr}{dt}[/tex] = ( -x / √( 625 - x² ) ) × 2
[tex]\frac{dr}{dt}[/tex] = -2x / √( 625 - x² )
∴
When x is 7 feet away from the wall;
[tex]\frac{dr}{dt}[/tex] = -2(7) / √( 625 - (7)² )
[tex]\frac{dr}{dt}[/tex] = -14 / √( 625 - 49 )
[tex]\frac{dr}{dt}[/tex] = -14 / √576
[tex]\frac{dr}{dt}[/tex] = -14 / 24
we simplify
[tex]\frac{dr}{dt}[/tex] = -7/12 ft/s
Therefore, When x is 7 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -7/12 ft/s
When x is 15 feet away from the wall
[tex]\frac{dr}{dt}[/tex] = -2(15) / √( 625 - (15)² )
[tex]\frac{dr}{dt}[/tex] = -30 / √( 625 - 225 )
[tex]\frac{dr}{dt}[/tex] = -30 / √400
[tex]\frac{dr}{dt}[/tex] = -30 / 20
we simplify
[tex]\frac{dr}{dt}[/tex] = -3/2 ft/s
Therefore, When x is 15 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -3/2 ft/s
When x is 20 feet away from the wall
[tex]\frac{dr}{dt}[/tex] = -2(20) / √( 625 - (20)² )
[tex]\frac{dr}{dt}[/tex] = -40 / √( 625 - 400 )
[tex]\frac{dr}{dt}[/tex] = -40 / √225
[tex]\frac{dr}{dt}[/tex] = -40 / 15
we simplify
[tex]\frac{dr}{dt}[/tex] = -8/3 ft/s
Therefore, When x is 20 feet away from the wall, [tex]\frac{dr}{dt}[/tex] = -8/3 ft/s
The Chernobyl reactor accident in what is now Ukraine was the worst nuclear disaster of all time. Fission products from the reactor core spread over a wide area. The primary radiation exposure to people in western Europe was due to the short-lived (half-life 8.0 days) isotope 131I
131 I, which fell across the landscape and was ingested by grazing cows that concentrated the isotope in their milk. Farmers couldn't sell the contaminated milk, so many opted to use the milk to make cheese, aging it until the radiactivity decayed to acceptable levels. How much time must elapse for the activity of a block of cheese containing 131I 131 I to drop to 1.0% of its initial value?
Answer:
The correct answer is "53.15 days".
Explanation:
Given that:
Half life of [tex]131_{I}[/tex],
[tex]T_{\frac{1}{2} }= 8 \ days[/tex]
Let the initial activity be "[tex]R_o[/tex]".and, activity to time t be "R".To find t when R will be "1%" of [tex]R_o[/tex], then
⇒ [tex]R=\frac{1}{100}R_o[/tex]
As we know,
⇒ [tex]R=R_o e^{-\lambda t}[/tex]
or,
∴ [tex]e^{\lambda t}=\frac{R_o}{R}[/tex]
By putting the values, we get
[tex]=\frac{R_o}{\frac{R}{100} }[/tex]
[tex]=100[/tex]
We know that,
Decay constant, [tex]\lambda = \frac{ln2}{T_{\frac{1}{2} }}[/tex]
hence,
⇒ [tex]\lambda t=ln100[/tex]
[tex]t=\frac{ln100}{\lambda}[/tex]
[tex]=\frac{ln100}{\frac{ln2}{8} }[/tex]
[tex]=53.15 \ days[/tex]
What happened to the combined energy of the two sleds when they collided?
Combined energy of the two sleds will be conserved when they collides.
What is Energy ?Energy is nothing but the ability to do work. there are different energies in different form which are thermal energy, mechanical energy, electric energy and sound energy etc.
According to first law of thermodynamic, Energy neither be created nor be destroyed. it can only be transferred from one form into another form. Energy is expressed in joule (J). its dimensions are [M¹ L² T⁻²].
Energy is conserved throughout the motion,
according to conservation law of energy, initial energy is equal to final energy.
When two sleds coming in opposite direction, it is having mass as well as velocity. hence it has kinetic energy, when they get collide with each other some of total energy gets converted into sound energy as collision cause sound(boom). some of the energy will use to break the material which made the sled and remaining energy will convert in kinetic energy of broken sleds it can be moved in opposite direction or in same direction depending initial energies.
in this collision combined kinetic energy can be converted into sound energy, mechanical energy and again kinetic energy but totally energy is conserved.
Hence combined energy will be conserved.
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When you blow across the top of a soda bottle, it acts like a closed pipe. If the inside of the bottle is 0.150 long, what is the fifth harmonic frequency
Answer:
The fifth harmonic frequency is 2.9 KHz.
Explanation:
For a closed tube, the fifth harmonic frequency is also the second overtone. So that;
fifth harmonic frequency = [tex]\frac{5V}{4L}[/tex]
where V is the velocity of the sound in air, and L is the length of the tube.
But L = 0.15 m and V = 343 m/s, then:
fifth harmonic frequency = [tex]\frac{5*343}{4*0.15}[/tex]
= [tex]\frac{1715}{0.6}[/tex]
= 2858.33 Hertz
fifth harmonic frequency = 2.9 KHz
The fifth harmonic frequency of the soda bottle is approximately 2.9 KHz.
Answer:
2828.33
Explanation:
answer for acellus
Hoover Dam on the Colorado River is the highest dam in the United States at 221 m, with an output of 1300 MW. The dam generates electricity with water taken from a depth of 100 m and an average flow rate of 640 m3/s
(a) Calculate the power in this flow in watts. 627200000XW Enter a number
(b) What is the ratio of this power to the facility's average of 680 MW? Additional Materials Reading
Answer:
a) 627840000 W
b) 0.9233
Explanation:
The rate at which energy flows through the dam for the generation of electricity is proportional to the volulme flow rate times the potential energy per unit volume.
Mathematically,
P = Q(ρgh)
Where, ρ is water density, g is accelaration due to gravity, h is height through which water falls over the turbine and Q is the discharge.
a) The power can be calculated as
P = 640(1000×9.81×221)
P = 627840000 W
b) The ratio = Calculated power / avarage power
= 62784×10^4/680×10^6
= 0.9233
a girl stands on a boat and feels the boat rise and fall 7 times In 1 minute. what is the period of the water wave? give your answer to 3 significant figures
Explanation:
hope it's helpful
have a nice day!!!! ;)
Which of the following is most consistent with the modern theory of evolution?
O Parents pass their physical traits to their offspring; those offspring with traits that help the
O Parents change their physical traits in order to survive in the environment, then those pa
O Life on this planet came from another planet far out in space.
O Living organisms have not changed for hundreds of millions of years.
Answer:
The theory of evolution by natural selection, first formulated in Darwin's book "On the Origin of Species" in 1859, is the process by which organisms change over time as a result of changes in heritable physical or behavioral traits
Please please help me please please help please please help me please
Answer:
I think its D, Does the wave travel In a Vaccum?
Answer:
It's C. Does the wave transfer energy?
Which of the following best explains the greater difficulty in stopping a 1000-kg car
moving at 174 km/h compared to an identical 1000-kg car moving at 100 km/h?
Stopping a 1000 kg car moving at a speed of 174 km/h will be more difficult than stopping an identical car moving at 100 km/h.
This can be explained by using Newton's second law of motion which says that the rate of change of momentum of a body is equal to the force applied.
We know that momentum of a body is the product of its mass and velocity.The momentum of the car moving at 174 km/h = [tex]1000\, kg \times 174\, km/h = 174000 \, kg.km/h[/tex].The momentum of the car moving at 100 km/h = [tex]1000\, kg \times 100\, km/h=100000 kg.km/h[/tex].Therefore, the car moving at 174 km/h has higher momentum and from Newton's second law of motion, a higher force would be required to stop this car.
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What was the name of the voyages taken by Zheng he during the Ming dynasty on behalf of China
Answer:
Ming treasure voyages
Explanation:
help ASAP 100 POINTS
Question 1 (1 point)
According to theory, which term best describes the universe before the big bang?
Question 1 options:
a star
a cloud of gas and dust
a small densely packed ball of matter
a planet
Question 2 (1 point)
Which example is used as evidence that the universe began with the big bang?
Question 2 options:
leftover energy in the form of background radiation in the universe from the event
the force of gravity
craters on the moon
the existence of atoms
Question 3 (1 point)
How do scientists use the Doppler effect to understand the universe?
Question 3 options:
to determine motion of objects
to determine light brightness
to determine planet compositition
to determine the ages of planets
Question 4 (1 point)
The Doppler effect indicates that the universe is expanding because_________.
Which phrase best completes the sentence.
Question 4 options:
light from other galaxies is the same color
light from other galaxies is the different color
light from other galaxies is blue shifted
light from other galaxies is red shifted
Question 5 (1 point)
What does a blue shift in light from stars indicate?
Question 5 options:
The stars are moving randomly.
The stars are moving closer.
The stars are moving farther away.
The stars are stationary.
Answer:
Explanation:
Question 1 (1 point)
According to theory, which term best describes the universe before the big bang?
a small densely packed ball of matter
Question 2 (1 point)
Which example is used as evidence that the universe began with the big bang?
Question 2 options:
leftover energy in the form of background radiation in the universe from the event
Question 3 (1 point)
How do scientists use the Doppler effect to understand the universe?
to determine motion of objects
Question 4 (1 point)
The Doppler effect indicates that the universe is expanding because_________.
Which phrase best completes the sentence.
light from other galaxies is red shifted
Question 5 (1 point)
What does a blue shift in light from stars indicate?
The stars are moving closer.
What drives waves?
Soda
Carbohydrates
Energy
Monochromatic light with lambda equals 630 space n m is incident on a single slit. The slit width is 0.8 mm. If the distance between the screen and slit is 1.8 m, what is the width of the central bright fringe on the screen
Answer:
The angular width is "[tex]1.575\times 10^{-3} \ rad[/tex]" and the linear width is "[tex]2.835\times 10^{-3} \ m[/tex]".
Explanation:
Given:
Wavelength,
[tex]\lambda = 630 \ nm[/tex]
or,
[tex]=630\times 10^{-9} \ m[/tex]
Slit width,
[tex]a = 0.8 \ mm[/tex]
or,
[tex]=0.8\times 10^{-3} \ m[/tex]
Distance between slit and screen,
[tex]D = 1.8 \ m[/tex]
As we know,
The central bright fringe's angular height,
⇒ [tex]\theta = \frac{2 \lambda}{a}[/tex]
[tex]= \frac{2\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]
[tex]=1.575\times 10^{-3} \ rad[/tex]
and,
The linear width will be:
⇒ [tex]x_o=\frac{2D \lambda}{a}[/tex]
By substituting the values, we get
[tex]=\frac{3\times 1.8\times 630\times 10^{-9}}{0.8\times 10^{-3}}[/tex]
[tex]=2.835\times 10^{-3} \ m[/tex]
Calculate the electric field intensity
at a Point 15cm from a charge of 10 uc
Answer:
it's very easy and simple answer u can't do it
The electric field intensity at a point of 15cm from a charge of 10 μc will be 3.9 ×10⁶ N/C.
What is the intensity of the electric field?
The electric field is defined as the area over an electric charge where its impact may be felt. The force experienced by a unit positive charge deposited at a given position is known as the electric field intensity.
The formula for the electric field intensity is given as;
[tex]\rm E= \frac{q}{4 \pi \epsilon d^2 } \\\\ \rm E= \frac{10 \times 10^{-6}}{4 \times 3.14 \times 8.85 \times 10^{-12}(15 \times 10^{-2}^2 } \\\\ E=3.9 \times 10^6 \ NC^{-1}[/tex]
Hence the electric field intensity at a point of 15cm from a charge of 10 μc will be 3.9 ×10⁶ N/C.
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A charged rod is brought near one end of a long, uncharged metal block. Students want to experimentally measure the resulting charge distribution along the entire length of the block. They have a small, positively charged sphere on a string that can be used as a test charge with negligible effect on the other charges. They will observe whether the sphere is attracted or repelled when held near the rod. Which of the following describes and justifies a procedure that will provide data to determine the entire charge distribution?
a. Hold the sphere near the end of the block closest to the rod, because that will give experimental data about both ends of the block.
b. Hold the sphere near each end of the block, because that will give experimental data about both ends of the block.
с. Hold the sphere near each end of the block and near the block's middle, because that will give experimental data about the area along the length of the block.
d. Hold the sphere near each end of the block and at a number of points along the length of the block, because that will give experimental data for the whole block.
Answer:
d
Explanation:
The sphere is positively charged so that when it comes in contact with the rod, its positive charges will repel to the other end. Hence, Hold the sphere near each end of the block, because that will give experimental data about both ends of the block.
What is polarization?Polarization is the separation of charges into two poles due electrostatic repulsion. Two similar charges will repel and opposite charges attracts each other.
When a charged or polar substance get in contact with a non-polar substance, the nonpolar one will deform in such a way that the electrons from the polar one will repel the electrons of nonpolar and there occurs a temporary charge separation in non polar substance creating an induced polarity.
Here, when the positive sphere comes in contact with the charged rod, the positive charges will align to the opposite pole of the rod away from the sphere and the negative charges will align near the sphere.
Thus there occurs an attractive force at one end and repulsion at the other end. Thus, holding the sphere at each end will data about charge distribution.
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what types of cuts do jig saw sanders make?
A column of argon is open at one end and closed at the other. The shortest length of such a column that will resonate with a 200 Hz tuning fork is 42.5 cm. The speed of sound in argon must be: Group of answer choices
Answer:
The speed of sound in the argon is 340 m/s.
Explanation:
Given;
fundamental frequency, f₀ = 200 Hz
length of the pipe, L = 42.5 cm = 0.425 m
A pipe that is open at one end and closed at another end is known as a closed pipe.
The wavelength for the first harmonic is calculated as;
L = Node -------> Antinode
L = λ/4
λ = 4L
The speed of the sound is calculated as;
v = fλ
where;
v is the speed of the sound
v = f x 4L
v = 200 x (4 x 0.425)
v = 340 m/s
Therefore, the speed of sound in the argon is 340 m/s.
A grindstone in the shape of a solid disk has a shaft attached to allow a force to be exerted on. The grindstone has a diameter of 0.650m and a mass of 55.0 kg. The shaft is 0.300 m from the center of the stone and has a mass of 4.00 kg. The grindstone has a motor attached and it is rotating at 450rev/min at a run when the motor is shut off. The grindstone comes to rest in 9.50 s.
a. What is the moment of inertia of the grindstone including the shaft?
b. What is the angular acceleration of the grindstone?
c. What average torque is done by friction to bring the grindstone to rest?
d. How many revolutions has it taken the grindstone to come to rest?
e. What is the linear acceleration of the shaft in this process?
f. What is the linear distance that the shaft traveled?
Answer:
a) 3.265 kg-m^2
b) - 4.96 rad/s^2
c) 16.1944 N-m
d) 35.625
e) - 1.488 m/s^2
f) 67.1175 m.
Explanation:
Given data:
Diameter of grindstone ( D ) = 0.650 m , Radius ( R ) = 0.325 m
mass of grindstone ( M ) = 55 kg
Radius of shaft ( r ) = 0.300 m
mass of shaft ( m ) = 4 kg
Initial Angular velocity = 450 rev/min = f = 7.5 rev/s = w =15π rad/s
time ( t ) = 9.50 secs
a) Determine the moment of inertia of the grindstone including the shaft
moment of inertia of grindstone = MR^2 / 2 = 55* (0.325)^2 / 2 = 2.905
moment of inertia of shaft = mr^2 = 4 *0.3^2 = 0.36
∴ moment of inertia including shaft = 2.905 + 0.36 = 3.265 kg-m^2
b) Determine the angular acceleration of the grindstone
∝ = - 15π / 9.5 ( i.e. angular velocity / time )
= - 4.96 rad/s^2 ( deceleration value )
c) Determine average torque done by friction to bring the grindstone to rest
Torque ( I * ∝ ) = 3.265 x 4.96 = 16.1944 N-m (magnitude)
d) Determine the number of revolutions before grindstone comes to rest
Total revolutions N before grindstone comes to rest
= ( f1 + f2)* t /2 = 7.5 * 9.5 / 2 = 35.625
Note : f2 = 0 as it comes to rest
f1 = 7.5 rev/s
e) Determine the Linear acceleration of the shaft in this process
This can be calculated using this relation
r * ∝ = 0.3 x (- 4.96 ) = - 1.488 m/s^2
f) Determine the linear distance travelled by the shaft
This can be calculated with the relation below
r * 2 * π * N = 0.3 * 2π * 35.625 = 67.1175 m.
what is the reason that causes viewing obstacle and less in sloar radiation?
Answer:
In addition to the sun's angle, atmospheric conditions can affect radiation levels. Cloud cover, air pollution and the hole in the ozone layer all alter the amount of solar radiation that can reach the surface. These factors all cause typical radiation levels to differ.
hope it helps
To solve this, we must be knowing each and every concept behind solar radiation. Therefore, cloud cover, air pollution, and the ozone hole all reduce the quantity of solar energy that reaches the earth's surface.
What is solar radiation?The energy generated by the sun as just a consequence of vast internal processes is known as solar radiation. In a word, the sun's capacity to generate intense nuclear fusion within and around the core enables it to radiate such vast amounts of energy inside the form of both heat and light. The entire process begins at the core of the sun.
Aside from the sun's angle, meteorological variables can influence radiation levels. Cloud cover, air pollution, and the ozone hole all reduce the quantity of solar energy that reaches the earth's surface. All of these variables contribute to variations in normal radiation levels.
Therefore, cloud cover, air pollution, and the ozone hole all reduce the quantity of solar energy that reaches the earth's surface.
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A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is
Each electron has a charge of :
[tex]1.6 \times 10 {}^{ - 19} \: C[/tex]
For producing 1 Cuolomb charge we need :
[tex] \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }[/tex][tex] \dfrac{10 {}^{19} }{1.6} [/tex][tex] \dfrac{10\times 10 {}^{19} }{16} [/tex][tex] \dfrac{100 \times 10 {}^{18} }{16} [/tex][tex] \mathrm{6.24 \times 10 {}^{18} \: \: electrons}[/tex]Now, for producing 0.009 C of charge, the number of electrons required is :
[tex]0.009 \times 6.24 \times {10}^{18} [/tex][tex]0.05616 \times 10 {}^{18} [/tex][tex] \mathrm{5.616 \times 10 {}^{16} \: \: electons}[/tex]_____________________________
So, Number of electrons passing through the cross section in 3.6 seconds is :
[tex] \mathrm{5.616 \times 10 {}^{16} \: \: electrons} [/tex]
Number of electrons passing through it in 1 Second is :
[tex] \dfrac{5.616 \times {10}^{16} }{3.6} [/tex][tex] \mathrm{1.56 \times 10 {}^{16} \: \: electrons}[/tex]Now, in 10 seconds the number of electrons passing through it is :
[tex]10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }[/tex][tex] \mathrm{1.56 \times 10 {}^{17} \: \: electrons}[/tex]_____________________________
[tex]\mathrm{ \#TeeNForeveR}[/tex]
A 16.2-g bullet with an initial speed of 850 m/s embeds itself in a 40.0-kg block, which is attached to a horizontal spring with a force constant of 1000 N/m. What is the maximum compression of the spring
The maximum compression in the spring is 0.06883 meters.
Given to us,
mass of bullet, [tex]m= 16.2\ grams[/tex]
velocity of bullet, [tex]v= 850\ meter/second[/tex]
mass of the block, [tex]m_2= 40\ kilograms[/tex]
velocity of block, [tex]v_2= 0\ meter/second[/tex]
spring constant, [tex]k = 1000\ Newton/meter[/tex]
Combined mass after collision,
[tex]\begin{aligned}M&= m_1+m_2\\&= 40.0162\ Kilograms\end{aligned}[/tex]
Using the momentum conservation equation, to find out final velocity of the system(bullet and block combined)
[tex]m_1v_1+m_2v_2= MV\\[/tex]
[tex]0.0162\times 850+40\times 0 = 40.0162\times V\\V= 0.3441 meter/sec[/tex]
Now using the Conservation of Energy (Kinetic energy of the system will be equal to potential energy in the spring), to find out the displacement of the spring [tex](x)[/tex]
[tex]\frac{1}{2}kx^2 = \frac{1}{2}MV^2\\\\\frac{1}{2}\times 1000\times x^2= \frac{1}{2} \times40.0162\times0.3441^2\\\\x^2= 0.004738\\\\x= 0.06883\ meters[/tex]
Hence, the maximum compression in the spring is 0.06883 meters.
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