Answer:
b. The internal resistance must be much smaller than the other resistances in the circuit.
Explanation:
Ammeter is used to measure the current flowing through a circuit. It is connected in series configuration with the load. In such a scenario the resistance of the ammeter should be negligible so as to make sure that the voltage drop across the resistance of ammeter is zero and it shows the correct reading of the current in the circuit.
Answer:
The average linear velocity (inches/second) of the golf club is 136.01 inches/second
Explanation:
Given;
length of the club, L = 29 inches
rotation angle, θ = 215⁰
time of motion, t = 0.8 s
The angular speed of the club is calculated as follows;
![\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%20%28%5Cfrac%7B%5Ctheta%7D%7B360%7D%20%5Ctimes%202%5Cpi%2C%20%5C%20rad%29%20%5Ctimes%20%5Cfrac%7B1%7D%7Bt%7D%20%5C%5C%5C%5C%5Comega%20%3D%20%20%28%5Cfrac%7B215%7D%7B360%7D%20%5Ctimes%202%5Cpi%2C%20%5C%20rad%29%20%5Ctimes%20%5Cfrac%7B1%7D%7B0.8%20%5C%20s%7D%20%5C%5C%5C%5C%5Comega%20%3D%204.69%20%5C%20rad%2Fs)
The average linear velocity (inches/second) of the golf club is calculated as;
v = ωr
v = 4.69 rad/s x 29 inches
v = 136.01 inches/second
Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second
Answer:Velocity = 6.325m/s
Directional angle= 18.43°
Explanation:
Using Right angle triangle
Let Velocity of ballon&hawk be VHB represent the height of the triangle.
Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.
Let Velocity of hawk and ground BE VHG represent the hypothesis.
Theta = opp/Adj= VHB/VBG
using pythagorean
VHG= SQRT(VHB^2+VBG^2)
VHG= sqrt(2^2+6^2)
VHG= sqrt(4+36)
VHG= 6.325m/s
Tan theta= 2/6
Tan theta =0.3333
Tan^-1 0.3333=18.43°
Answer:
As we are converting 220V AC into a 5V DC, first we need a step-down transformer to reduce such high voltage. Here we have used 9-0-9 1A step-down transformer, which convert 220V AC to 9V AC. In transformer there are primary and secondary coils which step up or step down the voltage according to the no of turn in the coils.
Selection of proper transformer is very important. Current rating depends upon the Current requirement of Load circuit (circuit which will use the generate DC). The voltage rating should be more than the required voltage. Means if we need 5V DC, transformer should at least have a rating of 7V, because voltage regulator IC 7805 at least need 2V more i.e. 7V to provide a 5V voltage.
the answer is c and if I help you thank me