Jeff goes to a nearby shop to buy some stationery. The figure below shows how his displacement s from his home changes with time t. The east is taken as positive. (a ) Where is the shop with respect to his home? (b) Describe Jeff's motion. (c) What are the total distance travelled by Jeff and his overall displacement?
Answers
A. The shop can be located either east or west depending on the direction of his displacement.
B. Jeff's motion can either be at a constant velocity, accelerating, or changing direction.
C. His overall displacement may be positive (or negative) and equal to the difference between his final and initial positions.
How did we arrive at these assertions?(a) To determine where the shop is located with respect to Jeff's home, you need to look at the direction of his displacement. If Jeff's displacement is in the positive direction (east), then the shop is located east of his home. If Jeff's displacement is in the negative direction (west), then the shop is located west of his home.
(b) Based on the information given about Jeff's displacement changing with time, you can infer some information about his motion. For example, if Jeff's displacement is increasing with time at a constant rate, then he is moving with a constant velocity. If Jeff's displacement is increasing with time but at a decreasing rate, then he is accelerating. If Jeff's displacement is increasing and then decreasing, then he is changing direction.
(c) The total distance travelled by Jeff is the sum of the absolute values of all his displacements, regardless of direction. Jeff's overall displacement is the vector sum of all his displacements, taking into account their direction. If Jeff ends up at the same location as his home, then his overall displacement is zero. If Jeff ends up at a location that is east (or west) of his home, then his overall displacement is positive (or negative) and equal to the difference between his final and initial positions.
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Related Questions
A runner covers the last straight stretch of a race in 8s.During that time he speeds up from 7 m/s to 9m/s. What is the acceleration
Answers
Given
Time taken is t=8 s.
The initial speed is u=7 m/s
The final speed is v=9 m/s.
To find
The acceleration
Explanation
We know the acceleration is the ratio of the difference in the speed to the time taken.
Thus,
\(\begin{gathered} a=\frac{v-u}{t} \\ \Rightarrow a=\frac{9-7}{8} \\ \Rightarrow a=\frac{2}{8}=\frac{1}{4}=0.25\text{ m/s}^2 \end{gathered}\)Conclusion
The acceleration is
\(0.25\text{ m/s}^2\)Consider an aircraft powered by a turbojet engine that has a pressure ratio of 12. The aircraft is stationary on the ground, held in position by its brakes. The ambient air is at 300 K and 95 kPa and enters the engine at a rate of 10 kg/s. The jet fuel has a heating value of 42,700 kJ/kg, and it is burned completely at a rate of 0. 2 kg/s. Neglecting the effect of the diffuser and disregarding the slight increase in mass at the engine exit as well as the inefficiencies of engine components, determine the force that must be applied on the brakes to hold the aircraft stationary, in kN
Answers
The force that must be applied on the brakes to hold the aircraft stationary is 84.51 kN (approx).
The force that must be applied on the brakes to hold the aircraft stationary is 84.51 kN.
Step 1:We can determine the exhaust gas temperature using the formulaT4 = T3 + (qin / (cp * m_dot)).Where,T3 = 1191.7 K (Given)qin = heating value of fuel × mass flow rate of fuel = 42,700 × 0.2 = 8,540 kJ/kgcp = specific heat at constant pressure = 1.005 kJ/kg.Km_dot = mass flow rate of air + mass flow rate of fuel = 10 + 0.2 = 10.2 kg/s
Putting all the values,T4 = 1191.7 + (8,540 / (1.005 * 10.2)) = 1191.7 + 84.85 = 1276.55 K
Step 2:The static thrust produced by the engine can be found out by the formula:
F = m_dot * (V2 - V1) + A2 * (P2 - P1)
Where,m_dot = mass flow rate of airA2 = area of the exhaust nozzleV1 = velocity of air at the engine inletV2 = velocity of air at the engine exitP1 = pressure of air at the engine inletP2 = pressure of air at the engine exitWe can use the simplified form of the above formula, which neglects the increase in the kinetic energy of the air in the combustor, the increase in the potential energy of the air from the inlet to the exhaust, and the change in the kinetic energy of the air from the inlet to the exhaust.F = m_dot * (Ve - V1) + Ae * (Pe - P1)
Where,Ve = velocity of air at the engine exitAe = area of the exhaust nozzlePe = pressure of air at the engine exitWe can calculate the value of Ve using the formula:Ve = sqrt(2 * cp * T4 * (1 - (P2 / P1)^((γ - 1) / γ))))Where,γ = ratio of specific heats = 1.4Putting all the values,Ve = sqrt(2 * 1.005 * 1276.55 * (1 - (1 / 12)^0.4))) = 724.23 m/sNow, we can calculate the static thrust,F = m_dot * (Ve - V1) + Ae * (Pe - P1)= 10.2 * (724.23 - 0) + (0.785 * (0.25)^2 * (12 * 10^3 - 95) * 10^3)= 74,044.47 N
Step 3:The force that must be applied on the brakes to hold the aircraft stationary is equal and opposite to the static thrust, that is, 74,044.47 N.
The value in kN will be: 74.04447/1000 = 74.04 kN.
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what is the magnitude of the force exerted by the biceps fbiceps ? what is the magnitude of the force exerted by the elbow felbow ? express your answers in newtons separated by a comma.
Answers
The force exerted by the biceps and the elbow is 239.9 N and 215.89 N respectively.
When we use only your forearms to lift an object, assume that your biceps are the main muscle that lifts your arm. Suppose the weight of the forearm he is 1.50 kg. If the biceps is attached to the forearm 2.50cm from the elbow, with the forearm parallel to the floor, flex the biceps to hold a 950g ball on the end of the elbow at a distance of 36.0cm from the elbow.
Against this background,
forearm mass = 1.50 kg
forearm length = 2.50 cm
The weight of the ball is 950 g.
36.0 cm ball spacing
Forearm strength needs to be calculated. Balance Elbow Using
Torque,
Fb x Db = Wf x Df/2 + Wball x Dball
Putting Value,
Fb x .025 = 1.5 x 9.8 x .36/2 + 0
95 x 9.8 x 0.36
Fb = 239.9 N
Now elbow force,
Fb = Fl + Wf + Wb
Fl = Fb - Wf - Wb
Putting value,
Fl = 239 .9 - (1.5 x 9.8 ) - (0.95×9.8)
Fl = 215.89 N.
Therefore the force on the elbow is 215.89 N.
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i) Show that total energy of the body at points A, B and C during the fall is same. ii) Find the distance from A to B and final velocity of the ball just reach before C. mass =5 kg, total height (h)= 100m
Answers
The total energy of the body at evevry point is remained same due to the law of conservation of energy. Distance from A to B and final velocity of the ball just reach before C is 44.3 m/s.
d (distance) from A to B is = √2gh
In this case given are, g = 9.8 m/s² and h = 100m,
so here d = √(2⋅9.8⋅100) = 44.3m.
Final velocity ,v = √2gh
Here given are , v is the velocity, g is the acceleration due to gravity, and h is the height. In this case,
g = 9.8 m/s² ,h = 100m,
v = √(2⋅9.8⋅100)
= 44.3 m/s (final velocity)
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A cyclist maintains a constant velocity of
5 m/s headed away from point A. At some
initial time, the cyclist is 248 m from point A.
What will be his displacement from his
starting position after 49 s?
Answer in units of m.
Answers
Hi there!
We can use the equation:
d = x₀ + vt, where:
x₀ = initial distance from the reference point
v = velocity (m/s)
t = time (sec)
Plug in the given values:
d = 248 + 5(49)
d = 493m
A 100 litre open-topped tank is filled to the brim with salt water. The salt concentration in the water is 80 g/L. Fresh water then pours into the tank at the rate of 10 L/sec. Assume the fresh water mixes completely with the salt water. The excess water runs out over the top (also at the rate of 10 L/sec). 10 L/s (fresh water) 100 L
a) Find an equation that gives the amount of salt remaining in the tank after t seconds.
b) How much salt is left in the tank after one minute?
c) How much salt is left in the tank after 100 L of brine has flowed out over the top of the tank?
d) When will half of the salt in the tank have flowed out over the top of the tank?
e) When will the tank contain salt water at a concentration of 5 g/L?
Answers
For the 100 litre open-topped tank:
a) rate of salt entering and leaving the tank: dS/dt = 0 - 80 = -80 g/secb) amount of salt left after 1 minute is 3200 gc) salt left in the tank after 100 L of brine has flowed out is 0d) half of the salt in the tank will have flowed out after 50 secondse) the tank will contain salt water after 250 seconds.How to solve for a 100 litre open-topped tank?a) To find an equation that gives the amount of salt remaining in the tank after t seconds, consider the rate at which salt enters and leaves the tank.
The initial amount of salt in the tank is 100 L × 80 g/L = 8000 g.
Rate of salt entering the tank:
The fresh water entering the tank at 10 L/sec has a salt concentration of 0 g/L. Therefore, the rate of salt entering the tank is 0 g/sec.
Rate of salt leaving the tank:
The excess water running out over the top at 10 L/sec carries away salt with a concentration of 80 g/L. Therefore, the rate of salt leaving the tank is 80 g/sec.
Rate of change of salt in the tank:
The rate of change of salt in the tank is given by the difference between the rate of salt entering and leaving the tank:
dS/dt = 0 - 80 = -80 g/sec
b) To find how much salt is left in the tank after one minute (60 seconds), integrate the rate equation from t = 0 to t = 60:
∫dS = ∫(-80) dt
ΔS = -80t + C
Since the initial amount of salt is 8000 g (at t = 0), solve for the constant C:
8000 = -80(0) + C
C = 8000
Therefore, the equation for the amount of salt remaining in the tank after t seconds is:
S(t) = -80t + 8000
Substituting t = 60 seconds:
S(60) = -80(60) + 8000
S(60) = 3200 g
c) After 100 L of brine has flowed out over the top of the tank, the remaining volume of salt water in the tank is 100 L (initial volume) - 100 L (flowed out) = 0 L. Therefore, there is no salt left in the tank.
d) To find when half of the salt in the tank has flowed out over the top, set S(t) = 0.5 × initial amount of salt:
0.5 × 8000 = -80t + 8000
-80t = 8000 - 4000
-80t = 4000
t = -4000 / -80
t = 50 seconds
Therefore, half of the salt in the tank will have flowed out over the top after 50 seconds.
e) To find when the tank contains salt water at a concentration of 5 g/L, set S(t) = V(t) × 5, where V(t) is the volume of water in the tank at time t:
V(t) × 5 = -80t + 8000
V(t) = (-80t + 8000) / 5
V(t) = 100 L (initial volume) - 10 L/sec × t (rate of water leaving the tank):
100 - 10t = (-80t + 8000) / 5
Multiplying both sides by 5:
500 - 50t = -80t + 8000
30t = 7500
t = 250 seconds
Therefore, the tank will contain salt water at a concentration of 5 g/L after 250 seconds.
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Why was a 7. 2 magnitude earthquake with an epicenter in Northern
Mexico felt over 100 miles away in the southern part of California?
Select one:
- Waves generated by earthquakes cause the entire planet to shake a little.
- Waves generated by earthquakes cause an increase in sound amplitude.
- Waves generated by earthquakes carry energy through the earth's crust.
- Waves generated by earthquakes carry sound through the earth's mantle.
Answers
The reason why a 7.2 magnitude earthquake with an epicenter in Northern Mexico was felt over 100 miles away in the southern part of California is due to waves generated by earthquakes carrying energy through the earth's crust. Therefore, the correct option is option 3
When an earthquake occurs, the seismic waves created by the movement of the tectonic plates travel through the earth's crust, causing the ground to shake. These waves are capable of traveling long distances and can be felt far away from the epicenter. This is because the energy released by the earthquake is transmitted through the solid rock of the earth's crust. As the waves travel through the ground, they cause the surface to vibrate, which can be felt by people and animals.
Therefore, the reason why the earthquake was felt over 100 miles away in the southern part of California is because the seismic waves generated by the earthquake carried energy through the earth's crust which corresponds to option 3.
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Electric Field Hockey
In this activity, you will again explore the relationship between an electric field and charged particles in the field, but this time you’ll have a gaming challenge. To begin, open Electric Field Hockey.
Directions:
On the control bar, make sure that the Puck is Positive and the Field boxes are checked. Also, make sure that the Practice option is selected.
Your aim is to score goals by manipulating the black puck (test charge) into the blue-colored bracket (goal) on the right. Think smart and place positive source charges (red) and negative source charges (blue) in such a way that the black puck moves into the goal.
Note that when you place a red, positive source charge in the hockey field, a red arrow appears on the black puck (test charge) showing the force the positive charge exerts on the puck. Similarly, when you place a blue, negative source charge in the hockey field, a blue arrow appears on the black puck (test charge) showing the force the negative charge exerts on the puck.
Question 1
Part A
Place a red charge on the hockey field and click Play. In which direction do electric field lines point? In which direction does the black puck move? What conclusion do you draw from this movement?
Part B
Click Reset and then click Clear. Now, place a blue charge in the hockey field and click Play. In which direction do electric field lines point? In which direction does the black puck move? What conclusion do you draw from this movement?
Part C
Manipulate the mass of the puck by dragging the mass bar to the right to increase the mass and to the left to decrease it. What changes do you see in the speed of the puck? Which principle works behind this change?
Part D
In the same situation, what do you observe about the relationship between the speed of the black puck and its distance from the blue charge?
Question 2
You can make the puck travel in complex ways by placing a set of charges around on the field. So, here’s your game challenge: Arrange sources charge around to propel the puck from its starting position into the goal. That’s pretty easy for a straight shot; you just put a negative charge behind the goal. But what if there are barriers along the way? This is a real test of your physical understanding, including Newton’s laws of motion and electrostatic forces. Game’s on!
Part A
On the control bar, check the Trace and Field boxes. The game has three Difficulty levels. Start with Difficulty level one and arrange source charges to get the puck into the goal. Once you’ve made a score at any level, increase the Difficulty level. Take a screen capture of two of your most difficult goals and paste them here. At least one of these should be at Difficulty level 2 or 3.
Please Answer Questions with complete sentences.
Answers
Answer:
The positive charges point away from each other
Explanation:
Why?
Arrows point away from the positive charge and toward the
negative charge.
Explanation:
Suppose that you rubbed a balloon with a sample of animal fur such as a wool sweater or even your own hair. The balloon would likely become charged and its charge would exert a strange influence upon other objects in its vicinity. If some small bits of paper were placed upon a table and the balloon were brought near and held above the paper bits, then the presence of the charged balloon might create a sufficient attraction for the paper bits to raise them off the table. This influence - known as an electric force - occurs even when the charged balloon is held some distance away from the paper bits. The electric force is a non-contact force. Any charged object can exert this force upon other objects - both charged and uncharged objects. One goal of this unit of The Physics Classroom is to understand the nature of the electric force. In this part of Lesson 1, two simple and fundamental statements will be made and explained about the nature of the electric force.
Perhaps you have heard it said so many times that it sounds like a cliché.
Opposites attract. And likes repel.
These two fundamental principles of charge interactions will be used throughout the unit to explain the vast array of static electricity phenomena. As mentioned in the previous section of Lesson 1, there are two types of electrically charged objects - those that contain more protons than electrons and are said to be positively charged and those that contain less protons than electrons and are said to be negatively charged. These two types of electrical charges - positive and negative - are said to be opposite types of charge. And consistent with our fundamental principle of charge interaction, a positively charged object will attract a negatively charged object. Oppositely charged objects will exert an attractive influence upon each other. In contrast to the attractive force between two objects with opposite charges, two objects that are of like charge will repel each other. That is, a positively charged object will exert a repulsive force upon a second positively charged object. This repulsive force will push the two objects apart. Similarly, a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other.
The activity where to explore the relationships between the electricity field charged with particles and those of the hockey field
The positively change will attract to each other and will opposite to the negatively changed ones. Arrows point away from the positive charge and toward the negative charge.Learn more about the you will again explore the relationship.
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If you push on a small car that is stuck in the mud, why do you sink into the
ground as you push on the car?
Answers
As you're applying force to the car, the mud which is a sinking surface would but you in the same place as the car is. The car is stuck in the mud and sunk to point where it cannot be pulled out unless an outside force interferes. When you apply force into the car and try to push it out, you would be put in the same situation. You would be experiencing friction which is an opposing force, in this problem, a simple machine such as a pulley would be helpful as it can lift the car from the outside. The car would act as a load and someone or some type of machine can use force to pick the car out of the mud. But if you were to try to push it out of the mud, you would just sink as the force applied would only put you in the same situation as the car, overall resulting in both beings (objects - car, human - "you" POV) being stuck in the same conumdrum.
Answer:
See below
Explanation:
Gravity will pull you down as you stand in the mud.
Additionally , if you do not push exactly horizontally ......like you push UP some , the equal and opposite force will push YOU down into the mud.
This would be like picking up a weight while you are standing there
F = ma will be greater and you will get muddier !
Which effect has been useful (and successful) in the search for and identification of black holes in the universe
Answers
Answer:
Detection of x-rays from a binary star undergoing mass exchange, where mass of component star can be determined.
A spring has a force of 2.0N and an extension of 0.30m. What will the extension be if the force applied is 6 N.
Will mark as brainliest
Answers
Answer:
0.9 m
Explanation:
Hooke’s Law
\(\large\boxed{F = ke}\)
where:
F is the force in newtons (N).k is the 'spring constant' in newtons per metre (N/m).e is the extension in metres (m).Given values:
F = 2.0 Ne = 0.30 mSubstitute the given values into the formula to find k:
\(\implies 2.0=0.30k\)
\(\implies k=\dfrac{2.0}{0.30}\)
\(\implies k=\dfrac{20}{3}\; \sf N/m\)
To find the extension if the force applied is 6 N, substitute the found value of k and the given value of F into the formula and solve for e:
\(\implies 6=\dfrac{20}{3}e\)
\(\implies 18=20e\)
\(\implies e=\dfrac{18}{20}\)
\(\implies e=0.9\; \sf m\)
The free-body diagram of a crate is shown. What is the net force acting on the crate? 352 N to the left 176 N to the left 528 N to the right 440 N to the right
Answers
On January 22, 1943, the temperature in Spearfish, South Dakota, rose from −4°F to 45°F in just 2 minutes. What was the temperature change in Celsius degrees?
27.2 C°
40.1 C°
35.4 C°
18.7 C°
Answers
The four phases of matter include all of the following except _____.
a. water
b. gas
c. liquid
d. plasma
e. solid
Answers
The correct answer is option A. Water is not a state/ phase of matter
What are the states of matter?States are those phases that describe its physical attributes as well as how its constituent particles are arranged. Solid, liquid, gas, and plasma are the four main states of matter that can exist.
When something melts, it transitions from a solid to a liquid condition.
Water has a freezing point of 0°C.The melting point of flour is around 220 degrees Celsius, which is greater than that of salt and sugar. Sublimation is the process by which a solid transforms into a gaseous state, as in the case of camphor. Examples of changes in the states of matter include the melting of ice, the freezing of water, the burning of camphor, the melting of wax, and the production of precipitates.
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when the top of the electrode leads the welding end of the electrode, and the welding arc is pointing back toward the weld bead, the travel angle is called a(n) ____ angle.
Answers
According to the information we can infer that the travel angle is called a "drag" angle.
How is it call a travel angle?In welding terminology, the travel angle refers to the angle between the axis of the electrode and the workpiece surface. When the top of the electrode leads the welding end (meaning the electrode is inclined downward at an angle) and the welding arc is pointing back toward the weld bead, it is known as a "drag" angle.
In this configuration, the electrode is dragged along the surface of the workpiece, creating a trailing arc direction. This technique is commonly used in certain welding processes, such as shielded metal arc welding (SMAW) or stick welding, to control the heat input and the direction of the molten metal flow during the welding operation.
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a 0.140-kg baseball traveling 39.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 15.0 cm . What was the average force applied by the ball on the glove?
Answers
The average force acting on the baseball is 709.8 N. The result is obtained by using the Newton's second law.
What is Newton's second law?The Newton's second law states that "The acceleration is directly proportional to the net force acting on an object and inversely proportional to the object's mass." It can be expressed as
a = ∑F / m
Where
a = acceleration∑F = net forcem = object's massA 0.140 kg baseball travelling 39.0 m/s strikes the catcher's mitt. It then makes the ball to rest and recoils backward 15.0 cm. What was the average force applied by the ball on the glove?
We have:
Initial seed, v₁ = 39.0 m/s.Final speed, v₂ = 0.Displacement, d = 15.0 cm = 0.15 m.The baseball moves from a certain speed to rest. So, it has acceleration. We can get acceleration with the kinematic equation.
v₁² = v₂² - 2ad
39² = 0 - 2a × 0.15
1521 = - 0.3a
a = - 5070 m/s²
The (-) sign means the object is slowing down.
Using the Newton's second law, the average force is
a = ∑F / m
F = ma
F = 0.140 × 5070
F = 709.8 N
Hence, the average force applied by the ball on the glove is 709.8 Newton.
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2. Given what you know about the acceleration of Earth's gravity (g = 9.8 m/s2), is this number accurate?
accurate. If not explain why you think it is not accurate. Pleaseee help mee
Answers
Answer:
it is correct
Explanation:
Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate
. which element is present in the fluorescent light? explain in detail how you can use your data to decide which gas is present 7. which of the gases that you tested are present in air?
Answers
The element present in the fluorescent light is mercury. To determine which gas is present, you can conduct a gas test.
The element present in a fluorescent light is mercury. When electricity is passed through the mercury vapor, it produces ultraviolet light that causes the phosphor coating inside the bulb to glow, creating visible light.
To determine which gas is present, you can analyze the emission spectrum of the light and compare it to known spectra of various gases. As for the gases present in air, they mainly include nitrogen (78%), oxygen (21%), and trace amounts of other gases like argon, carbon dioxide, and neon.
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House current is 120 volts. If a light bulb runs a current of 0.5 amps, what the resistance of the bulb? (Please please help!)
Answers
Resistance = (voltage) divided by (current)
= (120 V) / (0.5 Amp)
= 240 ohms .
roseeldavis
2 hours ago
Physics
High School
answered
Please help me with this question soon.
Block A - 19.3 Kg/L
Block B - 0.64 Kg/L
Block C - 0.70 Kg/L
Block D - 0.917 Kg/L
Block E - 3.53 Kg/L
1) Which block has the greatest density?
Block A
Block B
Block C
Block D
Block E
2) Which block has the lowest density?
Block A
Block B
Block C
Block D
Block E
3) Submit your data table and identify the substance each block is made of.
Answers
Answer: Block A - 19.3
Block B - 0.64
Block C - 0.70
Block D - 0.917
Block E - 3.53
Explanation: i hope this helps you a little
What is the smallest speed that one must hit a golf ball to travel a horizontal distance of 280 m in the air on a level fairway?
Answers
The correct answer is that the smallest speed required to hit a golf ball a distance of 280 m on a level fairway is approximately 63.8 m/s.
To find the smallest speed required to hit a golf ball a distance of 280 m on a level fairway, we can use the fact that the range of a projectile launched at an angle θ is given by: \(R = (v*2/g) * sin(2θ)\). Where v is the initial velocity of the projectile, g is the acceleration due to gravity, and θ is the launch angle. For maximum range, the launch angle is 45 degrees. So, we can rewrite the above equation as:\(v = sqrt((R * g) / sin(2θ))\). Plugging in the values given, we have: R = 280 m, g = 9.8 m/s*2, θ = 45 degrees. \(v = sqrt((280 m * 9.8 m/s*2) / sin(90 degrees)) ≈ 63.8 m/s\). Therefore, the smallest speed required to hit a golf ball a distance of 280 m on a level fairway is approximately 63.8 m/s.
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a bicycle racer sprints at the end of a race to clinch a victory. the racer has an initial velocity of 11.5 m/s when he was 300 m away from the finish line. he accelerates at the rate of 0.500 m/s/s for 7.00 s; and then the racer continues at the final velocity to the finish line. (a) what is his final velocity?
Answers
As a result, the racer's final velocity is 13.00 m/s after accelerating at 0.500 m/s2 for 7.00 s.
What is velocity?Velocity is a vector representation of an object's or particle's displacement with respect to time. The meter per second (m/s) is the standard unit of velocity magnitude (also known as speed). Alternatively, velocity magnitude can be expressed in centimeters per second (cm/s). The direction of movement of the body or item is defined by velocity. Speed is fundamentally a scalar number. Velocity is, in essence, a vector quantity. It is the pace at which distance changes. It is the displacement rate of change.
Here,
The final velocity can be found using the equation of motion,
vf = vi + at,
where vf is the final velocity, vi is the initial velocity (11.5 m/s), a is the acceleration (0.500 m/s^2), and t is the time for which the acceleration was applied (7.00 s).
Substituting the values we get,
vf = 11.5 + 0.500 * 7.00 = 13.00 m/s.
So, the racer's final velocity is 13.00 m/s as he accelerates at the rate of 0.500 m/s2 for 7.00 s.
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What operating pressure in atm is needed by a reverse osmosis system? Express your answer in atmospheres. Fresh water can be squeezed out of seawater, through a semipermeable membrane, by using a piston to raise the pressure on the seawater side of the membrane to a value larger than the osmotic pressure. To achieve a sufficient flow of fresh water, the operating pressure is typically twice the osmotic pressure. The membrane has to be able to withstand this pressure difference and also not become clogged too rapidly with any sediment or microorganisms that are too small to be filtered out of the seawater. The density of seawater is 1030 kg/mº, and a typical salinity is 3.5% NaCl by weight. Note that 1 mol of NaCl dissociates to 2 mol of solute particles in solution. Assume that the temperature is 20°C.
Answers
To achieve a sufficient flow of fresh water in a reverse osmosis system, the operating pressure is typically twice the osmotic pressure. The pressure needed depends on the osmotic pressure, which can be calculated based on the density and salinity of seawater.
In a reverse osmosis system, fresh water is obtained from seawater by applying pressure to overcome the osmotic pressure. The osmotic pressure is the minimum pressure required to prevent the flow of solvent (water) across a semipermeable membrane due to the concentration difference between the two sides.
To calculate the osmotic pressure, we need to consider the density and salinity of seawater. Given that the density of seawater is 1030 kg/m³ and the salinity is 3.5% NaCl by weight, we can calculate the number of moles of solute particles (NaCl) in one cubic meter of seawater.
Since 1 mol of NaCl dissociates into 2 mol of solute particles, the moles of solute particles per cubic meter of seawater can be determined.
The osmotic pressure can then be calculated using the ideal gas law, where osmotic pressure is equal to the concentration (in mol/m³) multiplied by the gas constant (R) and the temperature (in Kelvin). Assuming a temperature of 20°C, the osmotic pressure can be calculated.
Finally, the operating pressure in atm is typically twice the osmotic pressure, as stated in the problem.
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Collision 1: Blue Cart Initially at Rest Set the initial blue cart velocity to 0 m/s. Set the mass values to different values. Run the simulation and record the mass and velocity values. Before Collision After Collision mRed = _________ kg mBlue = _________ kg Use mass and velocity values to complete the following momentum table. Before Collision After Collision ∆Momentum Red Cart ________kg x _______m/s = ____________kg•m/s ________kg x _______m/s = ____________kg•m/s Blue Cart ________kg x _______m/s = ____________kg•m/s ________kg x _______m/s = ____________kg•m/s
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Answer:
it is very hard but i am trying to help you.
Explanation:
A box is accelerating to the right on a conveyor belt at a rate of 6 m/s2. A free body diagram of the box is shown with the normal, weight, and applied forces labeled.
Based on the free-body-diagram, determine the mass of the box in kilograms.
Answers
Answer:
15 kg
Explanation:
Since, weight=147 N
we know,
Weight=mass × acc.due to gravity (g)
or,mass=weight/g
or,mass=147/9.8
:.mass=15 kg
The force constant of a massless spring is 27.0 n/m. a mass of 0.35 kg is oscillating in simple harmonic motion at the end of the spring with an amplitude of 0.45 m. the maximum speed of the mass is?
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By using kinetic energy terms, the maximum speed of the mass is 3.95 m/s.
This massless spring can be solved by using kinetic energy terms. Kinetic energy can be determined as
K = 1/2 . m . v² ................... ( 1 )
where K is kinetic energy, m is mass and v is velocity.
When the case is on vibration, then the kinetic energy can be defined as
K = 1/2 . k . A² .................. ( 2 )
where k is force constant, A is amplitude of wave.
By combining these 2 equations, we get
K = K
1/2 . m . v² = 1/2 . k . A²
v² = k . A² / m
v = √(k . A² / m)
From the question above, we know that :
k = 27 N/m
m = 0.35 kg
A = 0.45 m
By substituting all parameters, we can determine the maximum speed
v = √(k . A² / m)
v = √(27 . 0.45² / 0.35)
v = 3.95 m/s
Hence, the maximum speed of the mass is 3.95 m/s
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what is physical properties of matter
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Answer:
A quality of matter that is unrelated to a change in its chemical makeup is called a physical property.
Examples of physical properties of matter:
DensityColorHardnessMelting PointBoiling PointElectrical ConductivityPhysical properties are any qualities of matter that may be felt or seen without changing the sample's chemical composition.
Due to the wide range of features that make up physical attributes, they are further separated into intensive and extensive as well as isotropic and anisotropic categories.
Any characteristic that can be recognized and measured by sight, hearing, touch, smell, or other senses without evoking a chemical reaction is referred to as a physical property. Examples of physical traits are:
Color, Shape, Volume, Density, Calorific value, Boiling point, Viscosity, Pressure, Solubility, electrical current
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Two airplanes leave an airport at the same time. The velocity of the first airplane is 650 m/h at a heading of 16.9 ◦ . The velocity of the second is 590 m/h at a heading of 161◦ . How far apart are they after 1.8 h? Answer in units of m.
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Distance of train after 1.8 h is 485.099 m
Distance is a measurement of how far apart two things or points are, either numerically or occasionally qualitatively. Distance can refer to a physical length in physics or to an estimate based on other factors in ordinary language.
Distance of the first plane = 650 x 1.8 = 1170 m
Distance of the second plane = 590 x 1.8 = 1062m
The angle between 2 planes
161◦ - 16.9◦ = 144.1°
Using the law of cosine
d^2 = (1170)^2 + (1062)^2 - 2*1170*1062*cos 144.1°
d = 485.099 m
Hence, distance of train after 1.8 h is 485.099 m
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calculate the pressure exerted by a force of 42N on a surface of 3m length and 2m bredth.[will be marking brainliest and following right answer..]
Answers
Answer:
Pressure, P = 7 Pa
Explanation:
Given that,
Force exerted on the surface, F = 42 N
Length, l = 3 m
Width, b = 2 m
We need to find the pressure exerted by this force. We know that, pressure is equal to force per unit area. So,
\(P=\dfrac{F}{l\times b}\\\\P=\dfrac{42}{3\times 2}\\\\P=7\ Pa\)
So, the required pressure is equal to 7 Pa.
what is an example of conservation
Answers
Answer:Talking to someone
Explanation: