Answer:
when pure water vapor from the water bodies get absorbed into polluted gases in the atmosphere, acid rain is caused and when this happens in a place where an element of the hydrosphere is located, it is polluted.in such a way hydrosphere is effected
Answer: When a parcel of air in the atmosphere becomes saturated with water, precipitation, such as rain or snow, can fall to Earth's surface.
Explanation:
Human contributions to greenhouse gases in the atmosphere are warming the earth's surface a process which is projected to increase evaporation of surface water and accelerate the hydrologic cycle. In turn, a warmer atmosphere can hold more water vapor.
12x+13y=12.01
x+y=2.323 times 10 to the 27 power
x= # of carbon - 12 atoms
y= # of carbon- 13 atoms
solve for x and y
Please help!!!!!
Answer:
[tex]y = - 2.7876 * 10^{28}[/tex]
[tex]x= 3.0199 * 10^{28}[/tex]
Explanation:
Given
[tex]12x+13y=12.01[/tex]
[tex]x+y=2.323 * 10^{27}[/tex]
Required:
Solve
Make x the subject in [tex]x+y=2.323 * 10^{27}[/tex]
[tex]x=2.323 * 10^{27} - y[/tex]
Substitute the above expression for y in [tex]12x+13y=12.01[/tex]
[tex]12(2.323 * 10^{27} - y) + 13y = 12.01[/tex]
Open brackets
[tex]12*2.323 * 10^{27} - 12y + 13y = 12.01[/tex]
[tex]12*2.323 * 10^{27} + y = 12.01[/tex]
Collect Like Terms
[tex]y = 12.01 - 12*2.323 * 10^{27}[/tex]
[tex]y = 12.01 - 2.7876 * 10^{28}[/tex]
12.01 is negligible compared to [tex]2.7876 * 10^{28}[/tex]
So:
[tex]y = - 2.7876 * 10^{28}[/tex]
Substitute the above expression for y in [tex]x=2.323 * 10^{27} - y[/tex]
[tex]x=2.323 * 10^{27} - (- 2.7876 * 10^{28})[/tex]
[tex]x=2.323 * 10^{27} + 2.7876 * 10^{28}[/tex]
[tex]x= 0.2323 * 10^{28} + 2.7876 * 10^{28}[/tex]
[tex]x= 3.0199 * 10^{28}[/tex]
How will the motion of the arrow change after it leaves the bow?
The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
pls thank me and brainliest me
Which of the following remains constant during the motion of a projectile fired from a planet?
A. Momentum
B. Kinetic energy
C. Horizontal component of velocity
D. Vertical component of velocity
Answer:
C. Horizontal component of velocity
Explanation:
Object in motion stays in motion,
nothing works against its motion in the horizontal direction, unlike in the vertical direction, gravity pulls object down.
The world’s largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds.
What is the velocity for one of the blades?
(THIS IS PHYSICS, CIRCULAR MOTION)
Answer:
88.14 m/s
Explanation:
From the question given above, the following data were obtained:
Radius (r) = length of blade = 80 m
Revolution (rev) = 1
Time (t) = 5.7 s
Velocity (v) =?
The velocity of the blade can be obtained by using the following formula:
v = (rev × 2πr) / t
NOTE: Pi (π) = 3.14
v = (1 × 2 × 3.14 × 80) / 5.7
v = 502.4 / 5.7
v = 88.14 m/s
Therefore, the velocity of the blade is 88.14 m/s
A system consists of two particles. Particle 1 with mass 2.0 k g 2.0kg is located at ( 2.0 m , 6.0 m ) (2.0m,6.0m) and has a velocity of ( 3.1 m / s , 2.6 m / s ) (3.1m/s,2.6m/s). Particle 2 with mass 4.5 k g 4.5kg is located at ( 4.0 m , 1.0 m ) (4.0m,1.0m) and has a velocity of ( 1.1 m / s , 0.6 m / s ) (1.1m/s,0.6m/s). Determine the position and the velocity of the center of mass of the system.
Answer:
[tex]Position:(3.38m ,2.53m)[/tex]
[tex]Velocity=(1.72m/s,1.22m/s)[/tex]
Explanation:
From the question we are told that
Mass of particle 1
[tex]M_1=2.0kg[/tex]
Co-ordinate of particle 1 (2.0m,6.0m)
Velocity of Particle 1
[tex]V_1= (3.1 m / s , 2.6 m / s )[/tex]
Mass of particle 2
[tex]M_2=4.5kg[/tex]
Co-ordinate of particle 2 (4.0m,1.0m)
Velocity of Particle 1
[tex]V_2=( 1.1 m / s , 0.6 m / s )[/tex]
Generally the Position is mathematically given as
[tex]X =\frac{ ((M_1 * Cx_1) + (M_2 *Cx_2)}{(M_1 + M_2)}[/tex]
[tex]X=\frac{2*2+4.5*4}{2+4.5}[/tex]
[tex]X=3.38[/tex]
[tex]Y=\frac{(M_1* Cy_1 + M_2* Cy_2)}{(M_1 + M_2)}[/tex]
[tex]Y =\frac{(2 * 6 + 4.5 * 1)}{(2 + 4.5)}[/tex]
[tex]Y= 2.53 m[/tex]
Therefore the position is given as
[tex]Position:(3.38m ,2.53m)[/tex]
Solving for Velocity
Generally the velocity of the system is mathematically Given as
[tex]V_x =\frac{ (M_1 * vx_1 + M_2 * Vx_2)}{(M_1 + M_2)}[/tex]
[tex]V_x=\frac{ (2 * 3.1 + 4.5 *1.1)}{(2 + 4.5)}[/tex]
[tex]V_x=1.72m/s[/tex]
For Y
[tex]V_y =\frac{(M_1* vy_1 + M_2* Vy_2)}{(M_1 + M_2)}[/tex]
[tex]V_y=\frac{ (2 * 2.6) + (4.5*0.6)}{(2 + 4.5)}[/tex]
[tex]V_y=1.22m/s[/tex]
Therefore Velocity
[tex]V=(1.72m/s,1.22m/s)[/tex]
The x-axis of a position-time graph represents
Answer:
The y-axis represents position relative to the starting point, and the x-axis represents time.
Explanation:
If a car travels 60 mph for a distance of 180 miles, how much time
did it take?
Answer:
3 hours
Explanation:
180 divided by 60 (mph means miles per hours by the way)
Why are stars considered to be the building blocks of the universe?
A. Stars are found throughout the galaxy
B. Stars become novas, supernovas, and Black Holes
C. Nuclear fusion within stars creates needed elements for he universe
D. The sun is a star
An adjustable tennis ball launcher launches tennis balls into the air from level ground and that return to level ground. The tennis balls are first launched with an initial velocity (vi) of 8.0 meters per second at an angle of 50° above the horizontal. The ball has an initial horizontal velocity of 5.1 meters per second. [Neglect friction.]
Calculate the vertical component of the ball’s initial velocity.
Calculate the maximum height reached by the ball.
Calculate the time elapsed to reach its maximum height.
Calculate the total horizontal distance travelled during its flight.
Answer:
Explanation:
I just need points
Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign are placed near each other.
Answer:
As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.
Explanation:
A circus performer walked up and to the right for a total displacement of 10\,\text m10m10, start text, m, end text along a diagonal tightrope angled 30 \degree30°30, degree above the ground.
What was the vertical displacement of the circus performer in \text mmstart text, m, end text?
Answer:
5m
Explanation:
khan academy
Vertical displacement of the circus performer is 5 m.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
A circus performer walked up and to the right for a total displacement of 10 m, start to end along a diagonal tightrope angled 30° degree above the ground.
Vertical displacement = 10 sin 30° = 5 m
Vertical displacement of the circus performer is 5 m.
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1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?
W = FdcosO =255 x 3 x cos 60 =
2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?
3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.
4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?
Answer:
(1)Work done by snail is 382.5 N
(2)Work done by ELEPHANT is 25160 N
(3)Work done by HANS is 2420 N
(4)Work done on block is 433 N
Explanation:
The work done due to force F applied at an angle θ from the horizontal to the body is give by
Work done = Fscosθ where, s is distance traveled by body
Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]
Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N
thus work done by snail is 382.5 N
Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]
Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N
thus work done by elephant is 25160 N
Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]
Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N
thus work done by Hans is 2420 N
Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]
Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N
thus work done on block is 433 N
(1) The work done by the sailor is 382.5 J.
(2) The work done by elephant is 25160 J.
(3) The work done by Hans upon the backpack is 2420 J.
(4) The required work done on block is 433 J.
Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,
[tex]W =F \times s \times cos \theta[/tex]
Here, s is the distance covered by the body.
(1)
Given data:
F= 255N , s= 3.00m and θ = 60.
The work done is calculated as,
W = Fscosθ
W= 255 x 3.00 x cos60 = 382.5 J.
Thus, the work done by the sailor is 382.5 J.
(2)
Given data:
F= 200N , s= 150m and θ = 33.
The work done by the elephant is calculated as,
W = Fscosθ
W= 200 x 150 x cos33. = 25160 J
Thus, the work done by elephant is 25160 J.
(3)
Given data:
F= 22.9N , s= 129m and θ = 35.
Then the work done upon the backpack is calculated as,
W= Fscosθ
W= 22.9 x 129 x cos35. = 2420 J
Thus, the work done by Hans upon the backpack is 2420 J.
(4)
Given data:
F= 100N , s= 5m and θ = 30.
The work done is calculated as,
W = Fscosθ
W= 100 x 5.00 x cos30. = 433 J
Thus, the required work done on block is 433 J.
Learn more about the work done here:
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A mechanism which uses mechanical energy to produce electrical energy is known as an) -
O electric motor
O transformer
O electromagnet
O electrical generator
what is the average gravitational force of attraction between the earth and the sun? the earth averages a distance of about 150 million km. the earth has a mass of 5.97x10^24 kg, and the sun has a mass of about 2x10^30 kg.
Answer:
B
Explanation:
Hhhhh
"When the ball leaves the ramp at Point B, students measure the horizontal" distance traveled. They repeat the experiment five times, being careful to release the ball from the same starting Point A and find the average horizontal distance traveled to be 2.0 m. One student suggests they use a stopwatch to find the time the ball is in the air whereas another student suggests they use a meter stick. Given these materials, describe the procedure students should follow to minimize error and calculate the speed of the ball as it leaves the ramp.
Answer:
Speed of ball as it leaves ramp is 2 m/s
Explanation:
From the question it is given that the average distance traveled by ball from point A to B is 2.0 m and the time taken is 1sec
one student is using stopwatch to calculate the time taken by ball and another student is using meter stick to calculate the distance traveled
since speed of ball is given by v = [tex]\frac{distance}{time}[/tex] thus,
the speed = 2 m/s
now for minimizing the error of speed it is necessary to record the readings by single students at-least 5 times and take average
by doing this, the error of speed calculation will be minimum as the it decreases the error due to random error of system caused by taking the reading by different students
“DNA...”
1: has a structure of a cone.
2:cannot be repaired if it is mutated.
3:is made up of nucleotides.
4:is made up of amino acids.
< Previous
Answer:
3. is made of nucleotides
How fast must a train accelerates from rest to cover 518 m in the first 7.48 s?
Heya!
For this problem, use the formula:
s = Vo * t + (at^2) / 2
Since the initial velocity is zero, the formula simplifies like this:
s = (at^2) / 2
Clear a:
2s = at^2
(2s) / t^2 = a
a = (2s) / t^2
Data:
s = Distance = 518 m
t = Time = 7,48 s
a = Aceleration = ¿?
Replace according formula:
a = (2*518 m) / (7,48 s)^2
Resolving:
a = 1036 m / 55,95 s^2
a = 23,34 m/s^2
The aceleration must be 23,34 meters per second squared
As an example, a 4.00-kg aluminum ball has an apparent mass of 2.10 kg when submerged in a particular liquid: calculate the density of the liquid.
Answer:
1.2825 * 10^3 kg/m³
Explanation:
Given that :
Mass of aluminum ball (m1) = 4kg
Apparent mass of ball (m2) = 2.10 kg
Density of aluminum (d1) = 2.7 * 10^3 kg/m³
Density of liquid (d2) =?
Using the relation :
d1 / d2 = m1 / (m2 - m1)
(2.7 * 10^3) / d2 = 4 / (4 - 2.10)
2700 / d2 = 4 / 1.9
4 * d2 = 2700 * 1.9
4 * d2 = 5130
d2 = 5130 / 4
d2 = 1282.5 kg/m³
Hence, density of liquid = 1.2825 * 10^3
The density of a solid or liquid material divided by the density of water is called
Answer:
I believe the answer is specific gravity
Explanation:Hope this helps :)
A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exert on the table if the brick is resting on its side?
Answer:
The pressure exerted by the brick on the table is 18,933.3 N/m².
Explanation:
Given;
height of the brick, h = 0.1 m
density of the brick, ρ = 19,300 kg/m³
acceleration due to gravity, g = 9.81 m/s²
The pressure exerted by the brick on the table is calculated as;
P = ρgh
P = (19,300)(9.81)(0.1)
P = 18,933.3 N/m²
Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².
Neon has 3 naturally occurring isotopes, Neon-20, Neon-21, and Neon-22. Neon's average atomic mass on the periodic table is 20.179 amu. Based on this information which isotope is most abundant? Explain why abundance of an isotope matters when calculating the average atomic mass of an isotope.
A LOT OF POINTS!! Best answer gets brainliest!
1) Which two of the following factors affect the momentum of an object: speed, velocity, mass, shape?
2) Can a small object can have a large momentum? Give an example.
3) How could you change the momentum of an object? What would you need to change?
4) A 5000kg truck and a 50 kg person have the same momentum. How is this possible?
how does a switch work in a circuit
An electric switch is a device that interrupts the electron flow in a circuit. Switches are primarily binary devices: either fully on or off and light switches have a simple design. When the switch is turned off, the circuit breaks and the power flow is interrupted. Circuits consist of a source of power and load.
A guitar player can change the frequency of a string by "bending" it-pushing it along a fret that is perpendicular to its length. This stretches the string, increasing its tension and its frequency. The B string on a guitar is 64 cm long and has a tension of 74 N. The guitarist pushes this string down against a fret located at the center of the string, which gives it a frequency of 494 Hz. He then bends the string, pushing with a force of 4.0 N so that it moves 8.0 mm along the fret.
* What is the new frequency?
Answer:
[tex]f'=504hz[/tex]
Explanation:
From the question we are told that
The B string on a guitar is 64 cm long
The B string tension tension of 74 N.
Frequency of 494 Hz
Pushed with a Force of 4.0 N
It moves 8.0 mm along the fret.
Generally the equation for frequency of ring under tension is mathematically given as
[tex]2Lf=\sqrt{x\frac{T}{\mu} }[/tex]
[tex]2*(64/100)*494=\sqrt{\frac{74}{m/0.64}[/tex]
[tex](632.32)^2={\frac{74}{m/0.64}[/tex]
[tex](632.32)^2=74*{\frac{0.64}{m}[/tex]
[tex](632.32)^2={\frac{47.36}{m}[/tex]
[tex]m=1.18450761*10^-^4[/tex]
Therefore finding the New frequency f'
[tex]f'=\frac{(\sqrt{\frac{74+11}{(\frac{1.18450761*10^-^4}{0.642})}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{74+11}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{85}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{54.57}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{678.7471973}{2*0.642}[/tex]
[tex]f'=572.9111096hz[/tex]
Assume that it takes an average of 3 man-hours to stack 1 ton of a particular item. In order to stack 36 tons in 6 hours, the number of people required is:______.
A) 9
B) 12
C) 15
D) 18
E) 21
Answer:
option (d)
Explanation:
Given,
1 ton of a particular item can be stack in= 3 man-hours
36 ton of a particular item can be stack in= 36 x 3 man-hours
= 108man- hours
Number of hours = 6 hours
Number of people = [tex]\frac{108 \text{man-hours}}{6}[/tex]
= 18
Hence, the number of people required is 18.
Therefore, option (d) is correct.
1. To get to school, a girl walks 1 km North in 15 minutes. She
then walks 200 m South-west in 160 seconds.
What is the girl's average velocity for her walk to school?
The girl's average velocity is 1.13 m/s
The formula used to determine average velocity is:
Velocity = distance/ timeV = D/tThis means to determine the girl's average velocity is necessary to find out the total distance and the total time.
Distance:
1 km north and 200 m SouthwestAs you can see there are two different distances and they are in different units (meters vs kilometers), let's convert the kilometers to meters so both distances are in the same unit.
1 x 1000 = 1000 (1 km = 1000 meters)Add the two distances:
1000 meters + 200 meters = 1200 meters
Time
15 minutes and 160 seconds
Let's convert the minutes into seconds:
15 x 60 = 900
Add the time:
900 + 160 = 1060
Finally, calculate the average velocity:
1200 meters / 1060 seconds = 1.13 meters per secondLearn more about velocity in: https://brainly.com/question/862972
A 0.290 kg block on a vertical spring with a spring constant of 5.00 ✕ 103 N/m is pushed downward, compressing the spring 0.110 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
___m
Answer:
The height at point of release is 10.20 m
Explanation:
Given:
Spring constant : K= 5 x 10 to the 3rd power n/m
compression x = 0.10 m
Mass of block m= 0.250 kg
Here spring potential energy converted into potential energy,
mgh = 1/2 kx to the 2 power
For finding at what height it rise,
0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power
h= 10.20
Therefore, the height at point of release is 10.20 m
Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/s2. How long does it take car A to overtake car B
Answer:
car A reaches and immediately overtakes the car B at 22.56 s.
Explanation:
After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:
[tex]x_{A}=300+x_{B}[/tex]
Now, we can rewrite this equation in terms of speed and time
[tex]V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t[/tex]
Where:
V(iA) is the initial speed of car A
V(iB) is the initial speed of car B
t is the time when car A reaches the car B
a is the acceleration
[tex]18t+\frac{1}{2}1.8t^{2}=300+25t[/tex]
[tex]0.9t^{2}-7t-300=0[/tex]
Solving this quadratic equation for t, and taking just the positive value, we will have:
t=22.56 s
Therefore, car A reaches and immediately overtakes the car B at 22.56 s.
I hope it helps you!
Unit Test
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TIME REMAININO
01:47:35
Which best describes the energy of a sound wave as it travels through a medium?
It increases
It decreases
It remains the same
It depends on the medium
Answer:
It depends on the medium is answer.
Explanation:
I hope it's helpful!
The energy of sound waves depends on the nature of medium it travels through. As the speed of the wave increases in the medium its energy increases. Hence, option d is correct.
What are sound waves ?Sound waves are type of mechanical waves passing through a medium. Sound waves are longitudinal waves hence, the oscillation of particles is in the same direction of of the wave propagation.
The energy of a wave is directly proportional to the frequency of the wave and inversely proportional to the wavelength. As the density of the medium increases, the speed of the wave decreases and thereby energy too.
Sound waves travels with higher speed through solids. As their energy increases, the waves moves with higher speed. Hence, the energy of sound waves depends on the nature of the medium.
Find more on sound waves:
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Which players are usually the fasted on the team, run the offensive, and shoot from the perimeter of the court?
Forward
Mid-fielder
Guard
Center
Answer:
i think guard but not sure