Answer:
A) If you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure b) 5 cm by 8 cm.
B) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure a) 3 cm × 5 cm
Explanation:
Resistivity is directly proportional to lenght and inversely properly to cross sectional area.
For the first case, 5 cm by 8 cm gives the largest area and leave 3 cm as the lenght. The resistivity of the metal will be smallest in these dimensions.
For the second case, 3 cm by 5 cm gives the smallest area, leaving 8 cm as the lenght. This is the maximum arrangement that can give the largest resistance possible.
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
Answer: amplitude
Explanation: This describes the maximum amount of the displacement of a particle from it rest position. Usually, it is measured in metres
Since we are considering AM which is amplitude modulation, a technique used in electronic communication, most commonly for broadcasting information through a radio carrier wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is diversified in proportion to that of the message signal being broadcasted.
Answer:A skier could benefit from a sports-specific training program because it could help them become better at skiing. ... Other activities such as squats and lunges will also improve skiing skills since it builds up lower body strength which is necessary to ski tough terrain.
Explanation:
Answer:at 21.6 min they were separated by 12 km
Explanation:
We can consider the next diagram
B2------15km/h------->Dock
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B1 at 20km/h
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V
So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.
Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.
