Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27
Simora [160]
Data D has the largest range.
Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23
Therefore, Data D has the largest range.
Answer:
The answer is B.
Explanation:
Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :
Total resistance = 20Ω + 40Ω + 60Ω
= 120Ω
Next, we have to find out that 1Ω is equal to how many voltage by dividing :
120Ω = 60V
1Ω = 60V ÷ 120
1Ω = 0.5V
Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :
1Ω = 0.5V
20Ω = 0.5V × 20
20Ω = 10V
The power that the light is able to utilize out of the supply is only 0.089 of the given.
Power utilized = (0.089)(22 W)
= 1.958 W
= 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J.
Answer:
Explanation:
The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:
where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes
T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:
w = mg and
5500 =- m(10) so
m = 550 kg. Now we have what we need to solve for the tension:
T = 550(4.0) + 5500 and
T = 2200 + 5500 so
T = 7700 N