In order to determine the required power, use the following formula:
[tex]P=\frac{W}{t}[/tex]where
t: time = 5.0 s
W: work
calculate the work W as follow:
[tex]W=F\cdot d=m\cdot a\cdot d[/tex]where
m: mass = 120 kg
a: acceleration = ?
d: distance
calculate the acceleration as follow:
[tex]a=\frac{v-v_0}{t}=\frac{10.0\text{ m/s- 5.0m/s}}{5.0s}=1\frac{m}{s^2}[/tex]the distance can be obtained by using the following formula:
[tex]d=v_0\cdot t+\frac{1}{2}at^2=(\frac{5.0m}{s})(5.0s)+\frac{1}{2}(\frac{1m}{s^2})(5.0s)^2=37.5m[/tex]then, replace the previous values of a and d to calculate W:
[tex]W=(120kg)(\frac{1m}{s^2})(37.5m)=4500J[/tex]Finally, replace W and d into the formual for the power P:
[tex]P=\frac{4500J}{5.0s}=900W[/tex]Hence, 900 watts are required to accelerate the car.
E. O-27.7 m/s9. An object of mass 2 kg moving with a speedof 26 m/s to the right collides with an objectof mass 13 kg moving with a speed of 2 m/sto the left. If the collision is completely inelastic,calculate their speed after collision. (1 point)A. O-0.831 m/sB. O 1.733 m/sC. 06.089 m/sD. O5.39 m/sE. O 0.388 m/s10. An object of mass 16 kg moving with a speed
Given
Mass of an object, m=2 kg
Velocity of the object, u=26 m/s towards right
Mass of the other object, m'=13 kg
Speed of the object, u'=2 m/s towards left
To find
If the collision is completely inelastic,calculate their speed after collision.
Explanation
Since the collision is perfectly inelastic, so after collision both the body would combine together and move as one
Let the velocity after collision be v.
Considering the direction towards right as positive.
Thus,
By conservation of momentum
[tex]\begin{gathered} mu+m^{\prime}u^{\prime}=(m+m^{\prime})v \\ \Rightarrow2\times26-13\times2=(2+13)v \\ \Rightarrow v=1.733\text{ m/s} \end{gathered}[/tex]Conclusion
The velocity is B. 1.733 m/s
27. Write a number in scientific notation with4 sig figs, and 2 zeros
Answer:
1.600 x 10³
Explanation:
The numbers in scientific notation are multiplied by a power of 10, additionally, if there is a decimal point, all the numbers including zeros are significant figures. Therefore, a number in scientific notation with 4 significant figures and 2 zeros is:
1.600 x 10³
Two forces act on an object as shown. Calculate the magnitude of the resultant force.[ Hint: Use the Pythagorean Theorem]
Using law of vector addition , the magnitude (R) of resultant force is given by
[tex]\begin{gathered} R=\sqrt{30^2+40^2+2\times30\times40\sin90\degree;} \\ R=\text{ }\sqrt{4900}=\text{ 70N}\begin{cases}sin90\degree={1} \\ \end{cases} \end{gathered}[/tex]Final answer is 70 N.
Note:- we need to use law of vector addition to get resultant of two forces but we cant use Pythagorean theorem for vector .
At the bottom of the first hill, the 300 kg car is traveling 39.6 m/s. What is the kinetic energy of the car at this point? (ignore friction forces)
ANSWER:
1st option: 235,000 J
STEP-BY-STEP EXPLANATION:
Given:
Mass (m) = 300 kg
Speed (v) = 39.6 m/s
We can calculate the kinetic energy using the following formula:
[tex]\begin{gathered} E_K=\frac{1}{2}mv^2 \\ \text{ We replacing:} \\ E_K=\frac{1}{2}\cdot300\cdot39.6^2 \\ E_K=235224\cong235000\text{ J} \end{gathered}[/tex]The kinetic energy is 235000 J.
Find the work W W done by the 30-newton force.
The work done by the 30 newton force is 4156.8 J
How do I determine the work done?We know that work done is defined according to the following formula:
Work done (Wd) = force (F) × distance (d)
Wd = fd
Where angle projection is involved, the work done is defined as follow::
Wd = FdCosθ
Haven know the formula for calculating work done, we shall thus, determine the work done by the 30 newton force as follow:
Force (F) = 30 newtonAngle (θ) = 30 degreesDistance (d) = 160 mWorkdone (Wd) =?Wd = FdCosθ
Wd = 30 × 160 × Cos 30
Wd = 4800 × 0.8660
Wd = 4156.8 J
Thus, from the calculation made above, we can conclude that the work done is 4156.8 J
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Missing part of question:
See attached photo
Find the final temperature of the thermometer assuming no heat flows to the surroundings
ANSWER
[tex]\begin{equation*} 71.51\degree C \end{equation*}[/tex]EXPLANATION
Parameters given:
Mass of thermometer, m = 300 g
Initial temperature of thermometer, t1 = 35°C
Volume of water, V = 258 cm³
Initial temperature of water, T1 = 80°C
First, let us find the mass of the water using the formula for density:
[tex]\begin{gathered} \rho=\frac{M}{V} \\ \Rightarrow M=\rho *V \end{gathered}[/tex]where ρ = density of water = 1 g/cm³
Therefore, the mass of the water is:
[tex]M=1*258=258g[/tex]According to the conservation of energy, the total heat flow (the sum of the heat energy of the thermometer and water) must be equal to 0 since no heat flows to the surroundings:
[tex]\begin{gathered} Q_g+Q_w=0 \\ mc(T-t_1)+MC(T-T_1)=0 \end{gathered}[/tex]where c = specific heat capacity of glass thermometer = 0.2 cal/g°C
C = specific heat capacity of water = 1 cal/g°C
T = final temperature of thermometer and water
Hence, solving for T, we have that:
[tex]\begin{gathered} T=\frac{mct_1+MCT_1}{mc+MC} \\ T=\frac{(300*0.2*35)+(258*1*80)}{(300*0.2)+(258*1)} \\ T=\frac{2100+20640}{60+258}=\frac{22740}{318} \\ T=71.51\degree C \end{gathered}[/tex]That is the final temperature of the thermometer.
a ray of light hits a flat mirror with an incident angle of 39.20 degrees. What angle does it reflect at? (do not put in units)
Given
A ray of light hits a flat mirror with an incident angle of 39.20 degrees.
To find
What is the angle of reflection?
Explanation
According to the law of reflection,
The angle of incidence is equal to the angle of reflection.
Here the angle of incidence is 39.20 deg
So the angle of reflection is 39.20 deg
Conclusion
The angle of reflection is 39.20 deg
a clay ball (0.65kg) is thrown at a wall and sticks. The speed of the ball before hitting the wall was 15 m/s. It took 45 miliseconds for the clay to come to a stop and make contact with the wall. What was the average force the wall exerted on the wall during the collision?
In order to determine the average force the wall exerted on the ball, use the following formula:
[tex]F=\frac{\Delta p}{\Delta t}[/tex]where
Δp: change in momentum
Δt: change in time = 45 ms = 0.045 s
F: force = ?
calculate the change in momentum as follow:
[tex]\Delta p=m\Delta v=(0.65kg)(15\frac{m}{s}-0\frac{m}{s})=9.75\text{ kg}\cdot\text{m/s}[/tex]next, replace Δp and Δt into the formula for F:
[tex]F=\frac{9.75\text{ kgm/s}}{0.045s}=216.67N[/tex]Hence, the average force exterded by the wall on the ball was approximately 216.67N
A 5.0n force is applied to a 3.0kg ball to change its velocity from +9.”m/s to +3.0 m/s. The impulse on the ball is _I think it’s -18
Given,
The force is F=5 N.
The mass is m=3.0 kg
The velocities are 9 m/s to 3 m/s
The impulse is:
[tex]\begin{gathered} I=m(v_f-v_i) \\ \Rightarrow I=3(9-3) \\ \Rightarrow I=\frac{18kgm}{s} \end{gathered}[/tex]The answer is 18 kgm/s
37.Why is hot water used to loosen a sealed jar lid?Select one:a. The jar contracts when heated.b. The lid expands when heated.c. The lid contracts when heated.d. The jar and the lid both contract.
Answer:
Explanation:
Heat causes expansion of the lid at a faster rate th
Part AA space vehicle accelerates uniformly from 80m/s att 0 to 167 m/s att 10.0 sHow far did it move between t = 2.0 s and t - 6.08 ?Express your answer to two significant figures and include the appropriate units.
Given:
Velocity at t = 0: 80 m/s
Velocity at t = 10.0 s: 167 m/s
Let's find the distance it covered between t = 2.0s and t = 6.0 s
Apply the kinematics formula:
[tex]v=u+at[/tex]Where:
v is the final velocity ==> 167 m/s
u is the initial velocity ==> 80 m/s
a is the acceleration
t is the time ==> 10.0 - 0 = 10.0 s
Now, let's find the acceleration.
Rewrite the formula for a:
[tex]\begin{gathered} a=\frac{v-u}{t} \\ \\ a=\frac{167-80}{10} \\ \\ a=\frac{87}{10} \\ \\ a=8.7m/s^2 \end{gathered}[/tex]The acceleration is 8.7 m/s²
To find the distance, apply the formula:
[tex]s=ut+\frac{1}{2}at^2[/tex]Thus, we have:
• At t = 2.0 s:
[tex]\begin{gathered} d_1=80(2)+\frac{1}{2}(8.7)(2)^2 \\ \\ d_1=160+17.4 \\ \\ d_1=177.4\text{ m} \end{gathered}[/tex]• At t = 6.0 s:
[tex]\begin{gathered} d_2=ut+\frac{1}{2}at^2 \\ \\ d_2=80(6)+\frac{1}{2}(8.7)(6)^2 \\ \\ d_2=480+156.6 \\ \\ d_2=636.6\text{ m} \end{gathered}[/tex]To find the distance traveled between t = 2.0 s and t = 6.0 s, we have:
d = d2 - d1 = 636.6 m - 177.4 m
d = 459.2 m
Therefore, the distance traveled betwen t = 2.0 s and t = 6.0 s is 459.2 m
ANSWER:
459.2 m
What will be the effect of multiplying negative scalar number with a vector.. Explain with the help of drawing.
When we multiply a vector by a scalar number, each dimension of the vector will be multiplied by the number.
For example, multiplying the vector <3, 4> by 2 would create the vector <6, 8>.
But if this scalar is negative, the dimensions of the vector will change the signal:
vector <3, 4> multiplied by -2 creates the vector <-6, -8>.
This causes the original vector to change its direction to the opposite direction, that is, it flips 180°:
MULTIPLE CHOICE QUESTION Which word is synonymous to tension? Push Pull
Pull
Explanation:Tension is a pulling force that is transmitted in a string, rope or cable.
Therefore, tension is synonymous to pull
The diagram below shows an example of an object on which tension is transmitted.
It is a pull because it is tranmitted towards the source rather than away from the source
What is the escape speed (in km/s) from an Earth-like planet with mass 4.9e+24 kg and radius 70.0 × 105 m? Use the gravitational constant G = 6.67 × 10-11 m3 kg-1 s-2.
The escape velocity = 9.66 km/s
Explanation:The mass of the planet is represented by m
[tex]m=4.9\times10^{24}\operatorname{kg}[/tex]The radius is represented by r
[tex]r=70.0\times10^5m[/tex]The gravitational constant is represented by G
[tex]G=6.67\times10^{-11}m^3kg^{-1}s^{-2}[/tex]The escape velocity (v) is given by the formula:
[tex]\begin{gathered} v=\sqrt[]{\frac{2GM}{r}} \\ v=\sqrt[]{\frac{2\times6.67\times10^{-11}\times4.9\times10^{24}}{70\times10^5}} \\ v=\sqrt[]{\frac{65.366\times10^{13}}{70\times10^5}} \\ v=\sqrt[]{0.9338\times10^8} \\ v=\sqrt[]{93380000} \\ v=9663.33\text{ m/s} \\ v=9.66\text{ km/s} \end{gathered}[/tex]The escape velocity = 9.66 km/s
Which of the following is NOT a characteristic of average speed?elloa. Average speed includes both the rate of motion and its direction.b. Average speed is the rate of motion.C. Average speed is always a positive value.d. The greater the speed of an object the faster it moves.
The average speed is defined as the total distance traveled by an object divided into the time that it takes for the object to travel that distance.
Average speed is a scalar quantity, which means that it can be represented using a number with no need of specifying a direction. Since the total distance is a positive value, then average speed is always also a positive value.
From the options, the statement "aveage speed includes both rate of motion and its direction" is false.
Therefore, the answer is: Option A.
The spectral classification for spica is B1V. This means it is A. A red giant star since B1 means it is very large B. A whit dwarf star since it is B1,hence it lies along the left edge of the HR for blue C.a blue giant star since it is B1, whee B stands for blue D. A main sequence star as indicated by V
The given spectral classification for spica is B1V.
B1 means it is B type star which have very high temperature and high mass. Due to its high temperature it will emit the light of lower wavelength which corresponds to blue color.
Thus, B1V is a white dwarf star since it is B1, hence it lies along the left edge of HR for blue which means option (B) is correct.
Which element is a good material for use as a semiconductor?copperoxygensiliconsodium
To find:
Which of the given elements is a good material for semiconductor?
Explanation:
The silicon is widely used for the manufacturing of semiconductors. The stable structure of the silicon makes it a suitable candidate for the manufacture of semiconductors. The silicon is widely abundant in nature. This makes it economical.
One closed organ pipe has a length of 2.24 meters. When a second pipe is played at the same time, a beat note with a frequency of 1.1 hertz is heard. By how much is the second pipe too long? Include units in your answer.
ANSWER
0.07 m
EXPLANATION
First, we have to find the frequency that the first pipe would play,
[tex]\lambda=4L=4\cdot2.24m=8.96m[/tex]The frequency is,
[tex]f=\frac{v}{\lambda}=\frac{343m/s}{8.96m}=38.28125Hz[/tex]The frequency played by the second pipe is,
[tex]f=38.25125Hz-1.1Hz=37.18125Hz[/tex]The wavelength of this note is,
[tex]\lambda=\frac{v}{f}=\frac{343m/s}{37.18125s^{-1}}\approx9.2251m[/tex]So the length of the second pipe is,
[tex]L=\frac{\lambda}{4}=\frac{9.2251m}{4}\approx2.31m[/tex]The difference between the pipes' length is,
[tex]2.31m-2.24m=0.07m[/tex]Hence, the second pipe is 0.07 meters too long
Consider a container filled with coconut oil having a density of 0.903 g/ml, What is the gauge pressure (Pa) 6.35 cm below the surface? 1ml = 1cm^3, 1000g = 1kg
The gauge pressure is 562.5 Pa.
Given data:
The density of oil is ρ=0.903 g/ml.
The distance below the surface is h=6.35 cm.
The density in kg/cm³ will be,
[tex]\begin{gathered} \rho=0.903\text{ g/ml }\times\frac{1\text{kg}}{1000\text{ g}}\times\frac{1\text{ml}}{1\text{ }cm^3}\times\frac{10^6\text{ }cm^3}{1m^3} \\ \rho=\frac{903kg}{m^3} \end{gathered}[/tex]The gauge pressure can be calculated as,
[tex]\begin{gathered} p=\rho gh \\ p=(\frac{903kg}{m^3})(\frac{9.81m}{s^2})(6.35cm\times\frac{1\text{ m}}{100cm}) \\ p=\frac{562.5kg}{ms^2}\times\frac{1\text{ Pa}}{\frac{kg}{ms^2}} \\ p=562.5\text{ Pa} \end{gathered}[/tex]Thus, the gauge pressure is 562.5 Pa.
During a hockey game, two hockey players (m1 = 82kg and m2 = 76kg) collide head on in a 1 dimensional perfectly instant collision. If the first hockey player is moving to the left at a velocity 4.2 m/s and the second hockey player is moving in the opposite direction at a velocity of 3.4 m/s, how fast are they both moving after they collide, assuming they stick together after they collide? How much kinetic energy is lost as a result of the collision?
Given:
The mass of player 1 is m1 = 82 kg
The initial velocity of player 1 is
[tex]v_{i1}=\text{ 4.2 m/s}[/tex]towards left.
The mass of player is m2 = 76 kg
The initial velocity of player 2 is
[tex]v_{i2}=\text{ -3.4 m/s}[/tex]in opposite direction.
Required:
The final velocity after the collision assuming they stick together.
The loss of kinetic energy after collision.
Explanation:
According to the conservation of momentum, the final velocity will be
[tex]\begin{gathered} m1v_{i1}+m2v_{i2}=(m1+m2)v_f \\ v_f=\frac{m1v_{i1}+m2v_{i2}}{m1+m2} \\ =\text{ }\frac{(82\times4.2)-(76\times3.4)}{82+76} \\ =0.544\text{ m/s} \end{gathered}[/tex]The loss in kinetic energy will be
[tex]\begin{gathered} \Delta KE=KE_{after}-KE_{before} \\ =\frac{1}{2}(m1+m2)(v_f)^2-\frac{1}{2}m1(v_{i1})^2-\frac{1}{2}m2(v_{i2})^2 \\ =\frac{1}{2}(82+76)(0.544)^2-\frac{1}{2}\times82\times(4.2)^2-\frac{1}{2}\times76\times(3.4)^2 \\ =23.38-723.24-439.28 \\ =-1139.14\text{ J} \end{gathered}[/tex]Final Answer:
The final velocity after the collision assuming they stick together is 0.544 m/s.
The loss of kinetic energy after the collision is 1139.14 J.
an ocean moves at 22.9 m/s and has a wavelength of 27.0. with what frequency do the waves pass you?
Which of the following is the type of force that pulls objects towards the center of motion?A. Inertia B. MomentumC. AccelerationD. Centripetal
When an object is in a circular motion, the acceleration responsible for changing the direction of the movement is known as the centripetal acceleration:
This acceleration points towards the center of the circular motion.
Using the second law of newton, we can find the centripetal force Fc:
[tex]F_c=m*a_c[/tex]Therefore the force that pulls objects towards the center of motion is the centripetal force.
Correct option: D.
a student who weighs 556 Newtons climbs the stairway (vertical height of 4.0 m) in 25 s (a) how much work is done (b) what is the power output of the student
Given data:
Weight of the student;
[tex]\begin{gathered} F=mg \\ =556\text{ N} \end{gathered}[/tex]Height;
[tex]h=4.0\text{ m}[/tex]Time;
[tex]t=25\text{ s}[/tex]Part (a)
The work done by the student is given as;
[tex]W=Fh[/tex]Substituting all known values,
[tex]\begin{gathered} W=(556\text{ N})\times(4.0\text{ m}) \\ =2224\text{ J} \end{gathered}[/tex]Therefore, the work done by the student is 2224 J.
Part (b)
The power is defined as the rate of doing work. Mathematically,
[tex]P=\frac{W}{t}[/tex]Substituting alll known values,
[tex]\begin{gathered} P=\frac{2224\text{ J}}{25\text{ s}} \\ =88.96\text{ W} \end{gathered}[/tex]Therefore, the power output of the student is 88.96 W.
Please answer the questions
Answer:
Where's the question? So that i can answer it!
i would like some help with this problem, i already tried to complete it on my own but would like to check my answer
Given:
Two circuits with multiple resistors
To find:
The net resistance of the circuits
Explanation:
For the first circuit
The equivalent series resistance of the 1 and 2 is,
[tex]\begin{gathered} R_1+R_2 \\ =20+16 \\ =36\text{ ohm} \end{gathered}[/tex]The equivalent series resistance of the 4 and 5 is,
[tex]\begin{gathered} 10+14 \\ =24\text{ ohm} \end{gathered}[/tex]The net resistance of the resistances is,
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{36}+\frac{1}{24}+\frac{1}{12} \\ \frac{1}{R}=\frac{11}{72} \\ R=\frac{72}{11}\text{ohm} \\ R=6.54\text{ ohm} \end{gathered}[/tex]Hence, the net resistance of the upper circuit is 6.54 ohms.
For the circuit below:
[tex]R_2,\text{ R}_3,\text{ R}_4\text{ are in series}[/tex]So, the equivalent resistance is,
[tex]\begin{gathered} R_2+R_3+R_4 \\ =2+2+2 \\ =6\text{ ohm} \end{gathered}[/tex]This equivalent resistance in parallel with
[tex]R_5[/tex]So, the equivalent resistance is,
[tex]\begin{gathered} \frac{6\times2}{6+2} \\ =1.5\text{ ohm} \end{gathered}[/tex]Now 1.5 ohm is in series with the rest two resistances.
So, the net resistance is,
[tex]\begin{gathered} R_1+R_6+1.5 \\ =2+2+1.5 \\ =5.5\text{ ohm} \end{gathered}[/tex]Hence, the net resistance of the circuit below is 5.5 ohms.
The zombies are closing in on him and Mr. Mangan sees a ramp to his left. Hedrives at it and hits the ramp with a velocity of 30 m/s. If it took him 7 s toget to the ramp, what was his acceleration?
Given data
*The given velocity is v = 30 m/s
*The given time is t = 7 s
The formula for the acceleration is given as
[tex]a=\frac{v}{t}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} a=\frac{30}{7} \\ =4.28m/s^2 \end{gathered}[/tex]Hence, the acceleration is a = 4.28 m/s^2
2. A Roborovski hamster attempts to cross a desert road in Mongolia. He first moves to the right 70 cm, then to the left 30 cm, the back to the right 90 cm and then finally back to the left for 370 cm. What is the total displacement of the hamster? What was the average speed of the little fellow if it took him 6's to bust those moves? What is the average velocity in the same time interval?
The total displacement of the hamster is 240 cm.
The average speed of the little fellow is 93.33 cm/seconds.
What is distance?Distance is a scalar quantity that indicates "how much ground an object has covered" while moving. Displacement is a vector quantity that describes "how far out of place an object is"; it is the overall change in the position of the object.
The average speed is calculated by dividing the total distance traveled by the total time it took to travel that distance. Speed is the rate at which something moves at any given time. Average speed is the average rate of speed over the course of a journey.
For total displacement:
taking right as +ve direction
Travel D = +70 - 30 + 90 - 370
Displacement = 240 cm
Total distance travelled will be:
= 70+30+90+370
= 560 cm
Average speed will be:
= Total distance travelled /time
= 560/6
= 93.33 cm/sec
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Christie and Daisy measured the applied forces while pulling on opposite ends of a rope, but they forgot to fill in their table ofresults shown below. Predict the most likely values that belong in Box C and Box D.Scale 1 (Christie) Scale 2 (Daisy) Observed Effect on Motion2 kg2 kgBox A3 kgBox BRope moved toward DaisyBox CBox DRope moved toward Christie(1 point)O Box C 4 kg; Box D: 4 kgO Box C: 3 kg, Box D: 3 kgO Box C: 4 kg, Box D: 3 kgO Box C: 3 kg; Box D: 4 kg
To find:
Most likely values that belong in box C and box D.
Explanation:
From Newton's second law of motion, the net force acting on an object is equal to the product of the mass of the object and the acceleration of the object produced due to the net force.
That is,
[tex]\vec{F}=m\vec{a}[/tex]Thus the object will accelerate in the direction of the net force.
It is given that in the third trail, the box moves towards Christie. Thus the net force is directed towards the Christie. Thus the force applied by Christie must be greater than the force applied by Daisy.
Final answer:
Thus the correct answer is option C.
The largest single publication in the world is the 1112-volume set of British Parliamentary Papers for 1968 through 1972. The complete set has a mass of 3,548 kg. Suppose the entire publication is placed on a cart that can move without friction. The cart is at rest, and a librarian is sitting on top of it, just having loaded the last volume. The librarian jumps off the cart with a horizontal velocity relative to the floor of 2.65 m/s to the right. The cart begins to roll to the left at a speed of 0.03 m/s. Assuming the cart’s mass is negligible, what is the librarian’s mass? Round to the hundredths.
ANSWER:
40.17 kg
STEP-BY-STEP EXPLANATION:
Given:
m1 = 3548 kg
V1 = -0.03 m/s
V2 = 2.65 m/s
We apply law of conservation of linear momentum:
[tex]\begin{gathered} P_i=P_f \\ P_i=0 \\ P_f=m_1\cdot V_1+m_2\cdot V_2 \end{gathered}[/tex]We replace and calculate the mass as follows:
[tex]\begin{gathered} m_1\cdot V_1+m_2\cdot V_2=0 \\ 3548\cdot-0.03+2.65\cdot m_2=0 \\ -106.44+2.65\cdot m_2=0 \\ m_2=\frac{106.44}{2.65} \\ m_2=40.17\text{ kg} \end{gathered}[/tex]Therefore the mass is equal to 40.17 kg
An ocean A. Open system B. Closed system C . Isolated
ANSWER:
A. Open system
STEP-BY-STEP EXPLANATION:
The ocean is a component of the hydrosphere, and the ocean surface represents the interface between the hydrosphere and the atmosphere above it. Which means that it generates an exchange of matter and energy, so the ocean is an example of an open system.