Given
A voice that is audible cannot be easily heard.
To find
Whether the given statement is true or false
Explanation
The sounds which a person can hear is known as audible. So the voice that is audible is easily heard.
Conclusion
The given statement is false
2. A plastic container weighs 56.7 g. When cooked pasta is placed in the container, the total weight is 170.1 g. What is the weight of the pasta?
The weight of the pasta is 113.4g.
How to find the weight of the pasta?The weight of an object is defined as the force of gravity acting on it, and it can be calculated by multiplying the mass by the acceleration of gravity, w = mg.
The formula for calculating an object's mass based on its weight is Mass is equal to Weight divided by Gravity's Acceleration Convert the weight in pounds to the equivalent weight in Newtons. Mass is measured in Newtons in the formula for calculating mass based on weight.
Given,
The plastic container weighs 56.7 g
Total weight is 170.1 g
Solution:
The weight of the pasta = Total weight - Plastic container's weight
= 170.1 - 56.7
= 113.4g
Therefore,
The weight of the pasta is 113.4g.
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Jennifer hits a stationary 0.20-kg ball, and it leaves her racket at 40 m/s. Time-lapse photography shows that the ball was in contact with the racket for 40 ms.What average force did the ball exert on the racket?Express your answer using two significant figures.What is the ratio of this force to the weight of the ball?Express your answer using two significant figures.
Given:
The mass of the stationary ball is: m = 0.20 kg.
The velocity of the ball when it leaves the rocket is: Vf = 40 m/s.
The time for which the ball was in contact with the rocket is: t = 40 ms = 0.04 s.
To find:
The average force the ball exerts on the rocket.
The ratio of the force on a rocket to the weight of the ball.
Explanation
As the ball is initially stationary, its initial velocity Vi is zero. After the ball was hit, it moves with the velocity Vf = 40 m/s for time t = 0.04 s.
Thus, using Newton's second law, we get:
[tex]\begin{gathered} F=ma \\ \\ F=m\times\frac{V_f-V_i}{t} \end{gathered}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} F=0.20\text{ kg}\times\frac{40\text{ m/s}-0\text{ m/s}}{0.04\text{ s}} \\ \\ F=200\text{ kg.}^\text{m/s}^2 \\ \\ F=200\text{ N} \end{gathered}[/tex]The weight of the ball on the earth is the product of its mass and acceleration due to gravity g. The value of the acceleration due to earth is: g = 9.8 m/s^2.
The weight W of the ball on earth is given as:
[tex]W=mg[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} W=0.20\text{ kg}\times9.8\text{ m/s}^2 \\ \\ W=1.96\text{ N} \end{gathered}[/tex]The ratio of the force F exerted by the ball on the rocket and the weight of the ball is:
[tex]\begin{gathered} \frac{F}{W}=\frac{200\text{ N}}{1.96\text{ N}} \\ \\ \frac{F}{W}=\frac{102.04}{1} \end{gathered}[/tex]Thus, F : W = 102.04 : 1.
Final answer:
The force exerted by the ball on the rocket is 200 N. The ratio of this force and the weight of the ball is 102.04 : 1.
Which sentence describes equilibrium in supply and demand? A golf dub manufacturing company decided to use a price-skimming strategy to enter the market. It set the price of its clubs at $450. However, a range of other premium clubs was already available in the market, and thus the product did not achieve a good demand. Eventually, when the price of the clubs fell to $300. demand for the product increased and profits began to soar. Now, market research suggests that if the company reduced its price to $275, the quantity demanded would also rise. The rise in demand was predicted to be 30 percent However, a profit study revealed that to increase demand, it would have to compromise on profits. Thus, the company decided to leave the price at $275. This would help them gain enough customers to make a profit while also manufacturing enough golf clubs to meet demand.
Given
A teenage boy was 4 feet 9 inches tall.
Since then, he has grown 9 inches.
To find:
The height of the boy.
Explanation:
It is given that,
A teenage boy was 4 feet 9 inches tall.
Since then, he has grown 9 inches.
That implies,
[tex]\begin{gathered} The\text{ }height\text{ }of\text{ }the\text{ }boy=4ft\text{ }9in+9in \\ =4ft\text{ }(9+9)in \\ =4ft\text{ }18in \\ =4ft\text{ }+1.5ft \\ =5ft\text{ }6in \end{gathered}[/tex]Hence, the height of the boy is 5feet 6inches.
It takes a snail 80 hours to travel 0.5 miles what is the speed of a snail
Answer:1/160
Explanation:
A series RLC circuit consisting of a capacitor of capacitive reactance
30
Ω
, a non-inductive resistor of
44
Ω
, and a coil of inductive reactance
90
Ω
and resistance
36
Ω
are connected across a 200-V, 60-Hz line as shown in the figure. Calculate the impedance of the circuit.
0.1
k
Ω
10.0
k
Ω
100.0
k
Ω
1.0
k
Ω
The impedance across the RLC circuit is 156.03 Ω
We are given that,
The resistance of the circuit = R = 36Ω
The voltage across the circuit = Vr = 200v
The capacitive reactance of the capacitor = χc = 30Ω
The non inductive- resistor of the circuit = χi = 90Ω
The maximum voltage across the circuit is given as,
V' = Iχc + Iχi + Vr
Where V' is the maximum voltage across the circuit, I is current, χc is capacitive reactance, χi is inductive reactance, Vr is voltage cross the resistor.
Thus, from Ohm's Law we can write as,
V = IR
I = V/R
I = (200v)/(36Ω)
I = 5.55amp
Therefore, the maximum voltage across the circuit can be calculated by putting all the values which is written above, thus, we get,
V' = Iχc + Iχi + Vr
V' = (5.55amp)(30Ω)+ (5.55amp)(90Ω)+200v
V' = 866.0 v
The impedance (Z) across the circuit is given by the maximum value of the voltage V' is divided by the maximum value of the current I , then the equation can be given as,
Z = V'/ I
Z = (866.0v)/ (5.55amp)
Z =156.03 Ω
Since, The impedance across the RLC circuit would be 156.03 Ω.
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please help me
how law of motion applied in tug of war games
Answer:
According to Newton's third law: For every action, there is an equal and opposite reaction. In a game of tug of war, the rope is the means to transfer forces. The first team to tug creates an action force, causing the pull on the rope, as the reaction force , the other teams rope end experiences a pull. If the second team initiates a pull, this is a second action force.
Professor Stauffer conducts basic research on the psychological, biological, and behavioral
factors that contribute to heart disease. She is most likely a
High blood pressure, high LDL cholesterol, diabetes, smoking, exposure to secondhand smoke, obesity, a poor diet, and inactivity are the main risk factors for heart disease and stroke.
Explain about the contribute to heart disease?High blood pressure, high cholesterol, and smoking are the three main risk factors for heart disease that at least half of all Americans (47%) have. Age and family history are two uncontrollable risk factors for heart disease.
The most frequent cause of coronary artery disease is atherosclerosis, which is a buildup of fatty plaques in the arteries. Poor diet, lack of exercise, obesity, and smoking are risk factors. Choosing a healthy lifestyle can help reduce the risk of atherosclerosis.
The most prevalent type of heart disease is coronary heart disease (CHD). It happens when plaque deposits cause the arteries that carry blood to the heart to constrict or stiffen. Fat, cholesterol, and other components found in blood make up plaque. The term "atherosclerosis" also refers to this plaque accumulation.
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Two coils , held in fixed positions , have a mutual inductance of 100 mH . What is the peak emf in one coil when the current in the other coil is i ( t ) = 10 sin ( 1000 t ) , where is in amperes and t is in seconds ?
The peak emf in one coil at the given current in the other coil is 1,000 A.
What is mutual inductance?Mutual Inductance between the two coils is defined as the property of the coil due to which it opposes the change of current in the other coil.
The peak emf in one coil when the current in the other coil is i ( t ) = 10 sin ( 1000 t ) depends on the mutual inductance and the peak current.
The peak emf in one coil is calculated as follows;
emf₀ = M di/dt
where;
M is the mutual inductancedi/dt is the rate of change of currentdi/dt = (10 x 1000) cos (1000 t)
di/dt = 10,000 cos(1000 t)
The peak rate of change of current from the derivative of di/dt = 10,000 A/s
emf₀ = (100 x 10⁻³) x (10,000 A/s)
emf₀ = 1,000 V
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What is the spring constant for a spring that stretches by 27 cm when a load of 200 N is suspended from it?1) 0.135 N/m2) 54 N/m3) 740 N/m4) 7.4 N/m
Given data:
Force;
[tex]F=200\text{ N}[/tex]Stretching;
[tex]\begin{gathered} x=27\text{ cm} \\ =0.27\text{ m} \end{gathered}[/tex]The force is given as,
[tex]F=kx[/tex]Here, k is the force constant.
The expression for the force constant is given as,
[tex]k=\frac{F}{x}[/tex]Substituting all known values,
[tex]\begin{gathered} k=\frac{200\text{ N}}{0.27\text{ m}} \\ =740.74\text{ N/m} \end{gathered}[/tex]Therefore, the force constant of the spring is 740.74 N/m. Hence, option (3); 740 N/m is the best choice.
A. 125 m
B. 27.5 m
C. 62.5 m
D. 55 m
(i don’t understand how to get the answer)
55m is the magnitude of the average distance of the car during this time interval.
The correct option is D.
How is distance measured in physics?The metre is the Si derived unit for distance (m). Long distances can be measured in kilometres, whereas little distances can be gauged in millimetres (cm) (km). For instance, you might measure the length in centimetres between the bottom and top of a piece of paper and the kilometres from your home and school.
What is Distance?The total movement of an object, independent of direction, is its distance. No matter where an object starts or ends, distance can be defined as the amount of space it has covered.
Briefing:The equation for the Total distance is mathematically given as
d=initial distance -final distance
d=90-35
d=55m
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The complete question is-
A car travels 90 meters due North in 10 seconds. The car turns around and travels 35 meters due south in 5 seconds. What is the magnitude of the average distance of the car during this time interval?
A. 125 m
B. 27.5 m
C. 62.5 m
D. 55 m
Answer:
27.5°
Subtract the summation of the angles from 180°
Explanation:
Before cutting the spring, the 800 kg car moves to the right with a speed of 12 m/s. After cutting the spring, determine the speed and momentum of the 900 kg car.
Given data:
* The mass of the car before cutting the spring is m_1 = 800 kg.
* The speed of the car before cutting the spring is u = 12 m/s.
* The mass of the car after cutting the spring is m_2 = 900 kg.
Solution:
According to the law of conservation of momentum, the momentum of the car remain conserved before and after cutting the spring,
[tex]m_1u=m_2v[/tex]where v is the final velocity of the car after cutting the spring,
Substituting the known values,
[tex]\begin{gathered} 800\times12=900\times v \\ v=\frac{800\times12}{900} \\ v=\frac{9600}{900} \\ v=10.66667\text{ m/s} \end{gathered}[/tex]Thus, the speed of a 900 kg car is approximately 10.67 m/s.
The momentum of 900 kg car is,
[tex]\begin{gathered} p=m_2v \\ p=900\times10.66667 \\ p\approx9600\text{ kgm/s} \end{gathered}[/tex]Thus, the momentum of the 900 kg car is 9600 kgm/s.
Identify the type of radioactive decay in the following reaction:a) Betab) Gammac) None of the aboved) Alpha
Given:
The chemical equation is shown in the question
To find:
The type of radioactive decay in the following reaction
Explanation:
In the case of Gamma emission, no mass number nor atomic number changes of daughter nuclei compared to mother nuclei.
In the case of alpha emission, mass number changes by 4 and atomic no by 2 of daughter nuclei compared to mother nuclei.
In the case of beta emission, only the atomic number of daughter nuclei changes, and it changes by 1.
Here, the daughter nuclei's atomic number is more than 1 compared to the mother nuclei.
Hence, the given emission is Beta.
Three-Step Conversion:Significant Figures are requiredHow many cm is 4.0 mi? (Remember your answer need to have the correct amount of significant figures and the correct units written).(1 mi = 5280 ft, 1 in=2.54 cm) (hint: cm -> in -> ft -> mi)
First, we need to transform 4mi to ft
we know
1mi=5280ft
[tex]4mi=4(5280)=21120ft[/tex]then we got 21120 ft to inches
1ft=12 in
[tex]21120ft=21120(12)=253440in[/tex]then we got 253440in to cm
1 in=2.54 cm
[tex]253440in=253440(2.54)=643737.6\text{ cm}[/tex]therefore the solution is
4.0mi= 643737.6 cm
One tugboat pulls on the barge with a force of magnitude 4000 units of force at 15° above the line AB (see the figure) and the other tugboat pulls on the barge with a force of magnitude 5000 units of force at 12° below the line AB. Resolve the pulling forces to their scalar components and find the components of the resultant force pulling on the barge.
The resultant components of force pulling on the barge is 8753.7 N.
What is the resultant force on the barge?
The resultant force on the barge is calculated by resolving the force into y and x-components.
The x and y-component of forces on the 4000 unit force is calculated as;
Fx = F cosθ
Fy = F sinθ
Fx = 4000 x cos(15)
Fx = 3863.7 N
Fy = 4000 x sin(15)
Fy = 1035.27 N
The x and y-component of forces on the 5000 unit force is calculated as;
Fx = 5000 x cos(12)
Fx = 4890.7 N
Fy = -5000 x sin(12)
Fy = -1039.6 N
Net horizontal force;
∑Fx = 3863.7 N + 4890.7 N
∑Fx = 8753.7 N
Net vertical force;
∑Fy = 1035.27 N - 1039.6 N
∑Fy = -4.33 N
The resultant force on the barge;
F = √[(-4.33²) + (8753.7²)]
F = 8753.7 N
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A cliff diver dove off a cliff at 8 m/s and landed in the water 24 m away from the cliff. How high is the cliff?
We are given the following information
Horizontal distance covered = 24 m
Initial speed = 8 m/s
We are asked to find the height of the cliff.
Let us first find the time.
[tex]\begin{gathered} v_{ix}=\frac{\Delta x}{t} \\ t=\frac{\Delta x}{v_{ix}} \end{gathered}[/tex]Where △x is the horizontal distance covered and vix is the initial horizontal speed of the diver.
[tex]t=\frac{24}{8}=3\; s[/tex]So, the time is 3 seconds.
The vertical distance covered by the diver is given by
[tex]y=\frac{1}{2}\cdot g\cdot t^2[/tex]Where g is the acceleration due to gravity that is 9.81 m/s^2
[tex]y=\frac{1}{2}\cdot9.81\cdot3^2=44.145\; m[/tex]Therefore, the height of the cliff is 44.145 m
where P = 12.5 and Q = 39.6.
What is the net force on the object? If the net force is to the right, enter a positive value and if it is to the left, enter a negative value.
The net force on an object will be equal to 27.1 N. The net force is in the right direction.
What is force?
Force can be described as the influence that changes the state of the body of motion or rest. The SI unit of force is Newton and force is a vector quantity. Force can change the direction as well as the speed of the moving object.
The force can be calculated from the product of the mass (m) and acceleration (a) of an object. The mathematical equation of the second law of motion for force is written as:
F = ma
The force P is acting in the left direction or pulling the object in the left direction. The force Q pulls the object in the right direction.
The net force [tex]=F_{P} -F_{Q}[/tex] = 12.5 - 39.6
The net force = 27.1 N in the right direction.
Therefore, the net force of 27.1 N on the object is pulling the object in the right direction.
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Your question is incomplete, most probably complete question was,
An object is pushed to the left with a force of P and pushed to the right with a force of Q. where P = 12.5 and Q = 39.6.
What is the net force on the object? If the net force is to the right, enter a positive value and if it is to the left, enter a negative value.
Your mission to Mars team needs a mars rover to slide down a frictionless ramp as shown to the right if the max acceleration that the rover can handle down the ramp before breaking apart is 0.513 m/s squared A. what is the maximum angle that the ramp can be made with the horizontal?B. Use your answer from A & the fact that Mars has a rover mass is 1025 kg to find the normal force on the inclined plane?
Given,
The maximum acceleration that the rover can handle, a=0.513 m/s²
(A)
The rover slides down the ramp under the influence of gravity. The component of the gravity that provides the rover with the required acceleration is the component that is along the ramp.
Thus,
[tex]g\sin \theta=a[/tex]Where g is the acceleration due to gravity and θ is the angle of inclination of the ramp.
On substituting the known values,
[tex]\begin{gathered} 9.8\times\sin \theta=0.513 \\ \Rightarrow\theta=\sin ^{-1}(\frac{0.513}{9.8}) \\ =3\degree \end{gathered}[/tex]Thus the maximum angle that the ramp can be made with the horizontal is 3°
(B)
Given, the mass of the rover, m=1025 kg
The normal force will be equal to the vertical component of the weight of the rover.
That is,
[tex]N=mg\cos \theta[/tex]On substituting the known values,
[tex]\begin{gathered} N=1025\times9.8\times\cos 3\degree^{} \\ =10031.2\text{ N} \end{gathered}[/tex]Thus the normal force on the rover is 10031.2 N
Two particles each with charge -Q are a fixed distance L apart, as show above. Each particle experiences a net electric force F. A particle with a charge +q is now fixed midway between the two particles. As a result the net electric force experienced by each negatively charged particle is reduced to F/2. What is the value of +q?QQ/2Q/4Q/8Q/16
The electric force between two charges can be calculated with the formula below (Coulomb's law):
[tex]F_e=\frac{K_e\cdot|q_1|\cdot|q_2|}{d^2}[/tex]Where Ke is the Coulomb's constant, q1 and q2 are the charges and d is the distance between them.
So, for Fe = F, q1 = q2 = Q and d = L, we have:
[tex]F=\frac{K_e\cdot Q^2}{L^2}[/tex]Now, after the addition of a positive charge in the middle of the charges, each negative charge will suffer another force, acting on the opposite direction of force F:
Since the new resulting force on the negative charges is F/2, the new force created by the positive charge addition is also F/2, so we have:
[tex]\begin{gathered} \frac{F}{2}=\frac{K_e\cdot Q\cdot q}{(L/2)^2} \\ F=\frac{2\cdot K_e\cdot Q\cdot q}{L^2/4} \\ F=\frac{8\cdot K_e\cdot Q\cdot q}{L^2} \\ \frac{K_e\cdot Q^2}{L^2}=\frac{8\cdot K_e\cdot Q\cdot q}{L^2} \\ Q=8q \\ q=\frac{Q}{8} \end{gathered}[/tex]Therefore the correct option is the fourth one: q = Q/8.
A 96.3 kg cannon at rest contains a 9.5 kg cannon ball. When fired, the cannon ball leaves the cannon with a speed of 84.58 m/s. What is the recoil speed of the cannon?
Using conservation of momentum:
[tex]p1=p2[/tex]Since the cannon is at rest before shots the ball:
[tex]\begin{gathered} p1=0 \\ _{\text{ }}and \\ p2=mv+MV \end{gathered}[/tex]Where:
[tex]\begin{gathered} m=9.5kg \\ v=84.58\frac{m}{s} \\ M=96.3kg \end{gathered}[/tex]So:
[tex]0=9.5\cdot84.58+96.3V[/tex]Solve for V:
[tex]\begin{gathered} 96.3V=-9.5\cdot84.58 \\ 96.3V=-803.51 \\ V=-\frac{803.51}{96.3} \\ V\approx-8.34\frac{m}{s} \end{gathered}[/tex]Answer:
The recoil speed of the cannon is approximately 8.34 m/s
7. A car traveling at 17.3 m/s starts to decelerate steadily. It comes to a complete
stop in 11 seconds. What is it's acceleration?
Answer:
-1.57 m/²
Explanation:
The equation is
v = u + at
where v = final velocity, u = initial velocity, a = acceleration, t = time
We can manipulate the final velocity equation to get
a = (v - u)/t
Final velocity v = 0, u = 17.3 m/s, t = 11s
a = (0 - 17.3)/11 = -17.3/11 = -1.57 m/²
Negative value means it is deceleration since speed is decreasing
HELP! Sketch the given vector and its horizontal and vertical components. Use trig ratios to find the magnitude of each component.
16 km 34 degrees N of W
Horizontal component = ?
Vertical component = ?
The x-component and the y-component of the given vector are -
[x] = - 13.36 a[x]
[y] = 8.947 a[y]
What is a vector quantity?A vector quantity is any physical quantity with both magnitude and direction. Examples - velocity, acceleration, displacement etc. Any vector quantity can be resolved into its components along the [x] and [y] axis as follows -
[x] component = AcosФ a[x]
[y] component = AsinФ a[y]
The resultant angle will be -
tanФ = |[y]|/|[x]|
Given is 16 km 34 degrees North of West
Now, in the image attached, it can be seen that a vector making an angle of 34 degrees is directed along the north of west. Now, the x-component will be -
[x] = 16 cos (34) -a[x] = - 13.36 a[x]
And, the y-component will be -
[y] = 16 sin (34) a[y] = 8.947 a[y]
Therefore, the x-component and the y-component of the given vector are -
[x] = - 13.36 a[x]
[y] = 8.947 a[y]
`
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Experiments and Data Collections and Interpreting Data
While studying paintball equipment, you may have learned that changes in temperature can cause paintballs to change size. At high temperatures, they swell, and at low temperatures, they shrink. This changes the way they fit in the marker barrel. You might affect their performance; if not, you can make an educated guess.
Include the following essential content:
* State a specific question about temperature and paintball performance.
Using the data provided, draw or create a simple graph that can be used to interpret the results.
Be sure to include these points:
*descriptive title
*dependent variable
*independent variable
*legend
At high temperatures, they swell, and at low temperatures, they shrink. This adjustment is the way they are in shape within the marker barrel. we may affect their overall performance.
Data Collection is the system of collecting and measuring statistics on focused variables in an established machine, which then allows one to reply to relevant questions and compare outcomes. statistics collection is a research issue in all study fields, which include physical and social sciences, humanities, and business.
Data Collection is the method of gathering and measuring facts on ation series variables of hobby, in a longtime systematic fashion that permits one to answer stated research questions, test hypotheses, and compare results.
Surveys, questionnaires and bureaucracy, completed online, in character or with the aid of smartphone, electronic mail or ordinary mail; awareness organizations and one-on-one interviews; and. direct observation of participants in a studies look at.
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In physics, a system is defined as the portion of the universe that has been chosen for study. Systems may be complex, such as a whole planet, or simple, such as the water inside a pot. The universe outside the system is called the surroundings. A closed system is one that does not exchange matter with its surroundings and does not experience any outside forces. An open system, on the other hand, can exchange matter with its surroundings and does experience outside forces.
The idea of systems and surroundings was not always part of physics. In 1824, the French physicist Nicolas Carnot applied the idea of systems to thermodynamics. A little later, German physicist Rudolf Clausius added the idea of surroundings. Today, system and surroundings are terms applied also to mechanical systems, as well as used in other scientific fields such as biology.
Examples of open systems are common in everyday life. A car driving down a road is a system. The car engine produces exhaust, which is matter passed to its surroundings. Friction between the car and the road is an outside force that acts on the system. Closed systems are rare in everyday life. They are commonly used in physics laboratories to simplify system analysis and calculations.
So, what questions can help someone decide whether the system under consideration is open or closed? The first question to ask is: Does the system exchange matter with its surroundings? If yes, it is an open system. If no, it is a closed system. The second question is: What is the source of the forces experienced by the system? If all forces originate within the system, the system is closed. If forces act on the system from outside of the system, it is an open system.
QUESTION 1
A student has chosen to study a swinging pendulum. The pendulum would be ____________.
OPTIONS
a system
the surroundings
a universe
a force
QUESTION 2
Which of the following statements describes a closed system?
OPTIONS
It exchanges matter with its surroundings.
Its surroundings act on the system.
It experiences only forces originating from within.
None of the above
QUESTION 3
Which of the following statements about cars indicates that they are an open system?
OPTIONS
Friction is present between different parts of the engine.
The car burns gasoline and produces exhaust.
The engine heats up as the car is driven.
The faster the car is driven, the faster its tires turn.
QUESTION 4
What is the most common use of closed systems?
OPTIONS
To minimize environmental damage
To create more efficient systems
To simplify system analysis
To predict system structure and behavior
QUESTION 5
Which of the following questions is important for deciding whether a system is open or closed?
OPTIONS
Does the system exchange matter with its surroundings?
Is the system composed of multiple parts?
Does the system contain more than one state of matter?
Is the system natural or man-made?
The pendulum would be a system. The statements describes a closed system is "It experiences only forces originating from within". The most common use of closed systems is to simplify system analysis
The pendulum would be a systemThe statements describes a closed system is the it experiences only forces originating from withinThe statement that indicates that the car is an open system is that the car burns gasoline and produces exhaustThe most common use of closed systems is to simplify system analysisTo decide whether a system is open or closed, the important question will be "Does the system exchange matter with its surroundings?"A closed system is one that does not exchange matter with its surroundings and does not experience any outside forces. An open system, on the other hand, can exchange matter with its surroundings and does experience outside forces
Therefore, the correct options are
A systemIt experiences only forces originating from withinThe car burns gasoline and produces exhaustTo simplify system analysisDoes the system exchange matter with its surroundings?To know more about open and closed system
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A 2 kg object and a 1 kg object are dropped from the same height.
Given that the inertia
masses of these objects are not the same, what hypothesis explains
why the two objects
fall with the same acceleration?
Select one:
O a. The force of gravity is the same for the 1 kg object as for the 2 kg object
O b. The force of friction is the same for the 1 kg object as for the 2 kg object.
O c. The force of gravity on the 1 kg object is twice as great as on the 2 kg object.
Od. The force of friction on the 2 kg object is twice as great as on the 1 kg object.
c. The force of gravity on the 1 kg object is twice as great as on the 2 kg object hypothesis explains why the two objects fall with the same acceleration.
Why do heavier and lighter objects fall at the same pace when dropped from the same height?If you disregard air resistance, things falling close to the surface of the Earth accelerate roughly at 9.8 m/s2 (9.8 m/s2, or g) because of gravity. As a result, the objects' acceleration is the same, and as a result, their velocity is likewise rising steadily.
No matter how heavy the object is, if there is no air resistance, the rate of descent simply relies on how far the object has dropped. This indicates that if two things are dropped simultaneously from the same height, they will both touch the ground at the same moment.
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If W = F x d, which of the following equations shows work being calculated using the correct units? (1 po
O 113 N= (17.4 J) x (6.51 ft)
O 113 J= (17.4 N) x (6.51 m)
O 113 N= (17.4 J) x (6.51 m)
O 113 J= (17.4 N) x (6.51 ft)
The equation that shows work being calculated using the correct units is as follows: 113J = (17.4 N) x (6.51 m) {option B}.
What is work done?Work done is a measure of energy expended in moving an object. It is most commonly calculated by multiplying force by distance. No work is done if the object does not move.
Work done is measured in Joules, which is equivalent to Newtons/metres (N/m).
Joules is the derived unit of energy, work and heat; the work required to exert a force of one newton for a distance of one metre.
Newton is the unit of measurement for force while metres is the unit of measurement for distance, hence, to calculate a work done of 113 Joules, 17.4 Newtons of force over a distance of 6.51 metres is needed.
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Answer:
It's 113 j = (17.4N) x (6.51)
Explanation:
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Which of the following describes how agricultural effects on forests differ from the oil industry’s effects on forests?Agricultural use leads to more damage from forest fires.Agricultural use does more harm to animal life in forests.Agricultural use requires the clearing of forest land for human’s needs.Agricultural use damages forest areas more slowly over a longer time period.
Agricultural use requires the clearing of forest land for human's needs.
Because oil damages the forest area over time due to pollution, agricultural use does not leads to more damage from forest fires and agricultural use does not harm animal life in forest. Therefore, third option is the correct one.
When the term quantized is use to refer to a charge it meansall charges are multiples of a protons chargethat the maximum value of a charge is 1.602x10-1⁹℃all charges are multiples of an electrons chargethat the minimum value of a charge is 1.602x10-1⁹℃O
all charges are multiples of an electrons charge.
Two objects collide and stick together. What will happen to the total kinetic energy of the objects (increase, decrease, stays the same)? Explain what happens to the kinetic energy.Effect on total kinetic energy: Description of Kinetic Energy changes:
Given:
Two objects collide and stick together.
To find:
Effect on the kinetic energy due to the collision.
Explanation:
When perfectly inelastic bodies moving along the same line collide, they stick together. An inelastic collision is a collision in which there is a loss in kinetic energy.
Let m1 and m2 be the masses and velocity v1 and v2 be the velocities of objects along the same line before the collision.
Let V be the velocity of objects after they collide and stick together.
By the law of conservation of linear momentum,
[tex]m_1v_1+m_2v_2=m_1V+m_2V[/tex]Rearranging the above equation, we get:
[tex]V=\frac{m_1v_1+m_2v_2}{m_1+m_2}\text{ ......\lparen1\rparen}[/tex]The total kinetic energy of objects before the collision is:
[tex]\text{Kinetic Energy \lparen before collision\rparen}=m_1v_1^2+m_2v_2^2\text{ ......\lparen2\rparen}[/tex]The kinetic energy after collision is:
[tex]\text{Kinetic energy \lparen after collision\rparen}=\frac{1}{2}(m_1+m_2)V^2\text{ ......\lparen3\rparen}[/tex]Using equations (1), (2), and (3), the loss in kinetic energy is given as:
[tex]\begin{gathered} \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2}\lbrack m_1v_1^2+m_2v_2^2-\frac{m_1v_1+m_2v_2}{m_1+m_2}\rbrack \\ \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{1}{2}\lbrack\frac{m_1m_2\left(v_1^2+v_2^2-2v_1v_2\right)}{m_1+m_2}] \\ \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{1}{2}(m_1+m_2)V^2=\frac{m_1m_2\left(v_1-v_2\right)^2}{2(m_1+m_2)} \end{gathered}[/tex]From the above result, we see that the energy loss in the kinetic energy is positive.
Thus, the kinetic energy of decreases, when two objects collide and stick together.
Final answer:
The kinetic energy of decreases, when two objects collide and stick together.
What will be it’s final speed and kinetic energy after it has been moving for t=5sec
Given data:
* The force applied on the block is,
[tex]F_a=12.5\text{ N}[/tex]* The frictional force acting on the block is,
[tex]F_r=4.2\text{ N}[/tex]* The acceleration of the block is,
[tex]a=1.2ms^{-2}[/tex]Solution:
The net force acting on the block is,
[tex]\begin{gathered} F_{\text{net}}=F_a-F_r \\ F_{\text{net}}=12.5-4.2 \\ F_{\text{net}}=8.3\text{ N} \end{gathered}[/tex]According to Newton's second law, the net force in terms of the mass and acceleration of the block is,
[tex]\begin{gathered} F_{\text{net}}=ma \\ m=\frac{F_{\text{net}}}{a}_{} \end{gathered}[/tex]Substituting the known values,
[tex]\begin{gathered} m=\frac{8.3}{1.2} \\ m=6.92\text{ kg} \end{gathered}[/tex]As the block is initially at rest, thus, the initial speed of the block is u = 0 m/s.
By the kinematics equation, the final speed of the block after t = 5 seconds is,
[tex]\begin{gathered} v-u=at \\ v-0=1.2\times5 \\ v=6\text{ m/s} \end{gathered}[/tex]Thus, the final speed of the block after t = 5 seconds is 6 meters per second.
The kinetic energy of the block after t = 5 seconds is,
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ K=\frac{1}{2}\times6.92\times6^2 \\ K=124.56\text{ J} \end{gathered}[/tex]Thus, the kinetic energy of the block after t = 5 seconds is 124.56 joule.
A small body A starts sliding down from the top of a wedge whose base is equal to l (see figure-23-4)The coefficient of friction between the body and the wedge surface is nue . At what value of the angle a will the time of sliding be the least?
At 49° value of the angle α, the time of sliding will be the least that is 1 seconds.
What is angle?When two straight lines or rays intersect at a single endpoint, an angle is created. The vertex of an angle is the location where two points come together. The Latin word "angulus," which means "corner," is where the word "angle" originates.
Let a be the acceleration of block when sliding and distance traveled by the block A from top of wedge to bottom of wedge.
d = 1 sec α
Friction force, f = kN = kmg cos α
Therefore, mg sin α − kmg cos α = ma ...(1)
Now, from kinematical equation: d = ut + (1/2)at²
[tex]$1 \sec \alpha=0+\left(\frac{1}{2}\right)at^2$[/tex]
[tex]$\mathrm{t}=\sqrt{\frac{21 \sec \alpha}{(\sin \alpha-\mathrm{k} \cos \alpha)}} \mathrm{g} \quad$[/tex] (using equation (1))
[tex]$$=\sqrt{\frac{21}{\left(\frac{\sin 2 \alpha}{2}-\mathrm{k} \cos ^2 \alpha\right)}} \mathrm{g} \ldots \ldots . .(2)[/tex]
[tex]$For $\mathrm{t}_{\min }, \frac{\mathrm{d}\left(\frac{\sin 2 \alpha}{2}-\mathrm{k} \cos ^2 \alpha\right)}{\mathrm{d} \alpha}=0$[/tex]
[tex]i.e. \frac{2 \cos 2 \alpha}{2}} +2 \mathrm{k} \cos \alpha \sin \alpha=0$[/tex]
[tex]$\cos 2 \alpha+\mathrm{k} \sin 2 \alpha=0$[/tex]
[tex]$\tan 2 \alpha=-\frac{1}{k} \Rightarrow \alpha=49^{\circ}$[/tex]
Putting the values of α, k and l in equation (2) we get [tex]t_{min}[/tex] = 1s
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Full question
A small body A starts sliding down from the top of a wedge as shown in figure above, whose base is equal to l=2.10m. The coefficient of friction between the body and the wedge surface is k=0.140. At what value of the angle α will the time of sliding be the least?
