Answer: A.AB
Explanation:
This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.
As we can see in the attached image, the graph can be divided in four segments:
OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.
AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>
BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.
CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.
From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:
Segment AB
As a system is cooled to its freezing point, the kinetic energy of the particles in the system will lower so the movement are much slower. Cooling means taking out heat from the system. This process is a physical change because it is only the phase of the system is changed and it is still the substance after the process.
Answer:20/47 meter per second
Explanation:
Mass of arrow(ma)=0.25kg
Velocity of arrow(va)=12m/s
Mass of target(mt)=6.8kg
Velocity of target(vt)=0 since target is at rest
Conservation of linear momentum says that :
maxva+mtxvt=(ma+mt)V
V=(maxva+mtxvt)/(ma+mt)
V=(0.25x12+6.8x0)/(0.25+6.8)
V=3/(7.05)
V=20/47 meter per second
Answer: Both Technician A and B are right.
Explanation:
Both Technicians are correct. when a valve remains open as it approaches the TDC(top dead center), they piston may strike the valve which would result to damage of the engine. Same applies when it is open as it approaches the top dead center as a valve train damage may occur. As the driven train is responsible for providing power to the wheels from the engine block.
'A' and 'C' are exactly the same circuit, except the voltmeter's terminals are flipped.
'A' is the correct way to hook everything up.
If you start at the positive terminal of the battery, and follow the flow of current through the circuit and around to the negative terminal, you're following the path where the voltage gets lower and lower and lower all the way.
So each time you come to any device in the circuit ... whether it's a resistor or a meter ... you would be hitting the positive side of it first, and then the voltage where you come out on the other side of it would be lower.
So the left side of the resistor is more positive, and the right side is more negative. The voltmeter is connected correctly in 'A', but it's backwards in 'C'. If you connect the voltmeter like in 'C' and turn things on, the voltmeter will try to go <em>down</em> from zero. You can't read the number on it, and It's possible that the voltmeter might be damaged.
