Answer:
It's called an ampere!
Explanation:
The SI unit of electric current is the ampere, which is the flow of electric charge across a surface at the rate of one coulomb per second. The ampere (symbol: A) is an SI base unit Electric current is measured using a device called an ammeter.
Hope this helps :)
Answer:
![N_s\approx41667 \hspace{3}lo ops](https://tex.z-dn.net/?f=N_s%5Capprox41667%20%5Chspace%7B3%7Dlo%20ops)
Explanation:
In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:
![\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings](https://tex.z-dn.net/?f=%5Cfrac%7BV_p%7D%7BV_s%7D%20%3D%5Cfrac%7BN_p%7D%7BN_s%7D%20%5C%5C%5C%5CWhere%3A%5C%5C%5C%5CV_p%3DPrimary%5Chspace%7B3%7D%20Voltage%5C%5CV_s%3DV_p%3DSecondary%5Chspace%7B3%7D%20Voltage%5C%5CN_p%3DNumber%5Chspace%7B3%7D%20of%5Chspace%7B3%7D%20Primary%5Chspace%7B3%7D%20Windings%5C%5CN_s%3DNumber%5Chspace%7B3%7D%20of%5Chspace%7B3%7D%20Secondary%5Chspace%7B3%7D%20Windings)
In this case:
![V_p=120V\\V_s=100kV=100000V\\N_p=50](https://tex.z-dn.net/?f=V_p%3D120V%5C%5CV_s%3D100kV%3D100000V%5C%5CN_p%3D50)
Therefore, using the previous equation and the data provided, let's solve for
:
![N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps](https://tex.z-dn.net/?f=N_s%3D%5Cfrac%7BN_p%20V_s%7D%7BV_p%7D%20%3D%5Cfrac%7B%2850%29%28100000%29%7D%7B120%7D%20%3D%5Cfrac%7B125000%7D%7B3%7D%20%5Capprox41667%5Chspace%7B3%7Dloo%20ps)
Hence, the number of loops in the secondary is approximately 41667.
Answer:
![\Delta P = -1.14 \hat i + 0.33\hat j](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20-1.14%20%5Chat%20i%20%2B%200.33%5Chat%20j)
Part b)
![v_f = 0.03 \hat i + 1.79 \hat j](https://tex.z-dn.net/?f=v_f%20%3D%200.03%20%5Chat%20i%20%2B%201.79%20%5Chat%20j)
Explanation:
Part a)
As we know that impulse is due to the force applied for small time due to which momentum is changed
so we will have
![\Delta P = F\Delta t](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20F%5CDelta%20t)
so we will have
![F = -380 \hat i + 110 \hat j](https://tex.z-dn.net/?f=F%20%3D%20-380%20%5Chat%20i%20%2B%20110%20%5Chat%20j)
also we know that
![\Delta t = 3 ms](https://tex.z-dn.net/?f=%5CDelta%20t%20%3D%203%20ms)
now impulse is given as
![\Delta P = (-380\hat i + 110 \hat j)(3 \times 10^{-3})](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%28-380%5Chat%20i%20%2B%20110%20%5Chat%20j%29%283%20%5Ctimes%2010%5E%7B-3%7D%29)
![\Delta P = -1.14 \hat i + 0.33\hat j](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20-1.14%20%5Chat%20i%20%2B%200.33%5Chat%20j)
Part b)
As we know that
![\Delta P = m(v_f - v_i)](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20m%28v_f%20-%20v_i%29)
![(-1.14 \hat i + 0.33 \hat j) = (\frac{0.560}{9.81})(v_f - (20\hat i - 4\hat j))](https://tex.z-dn.net/?f=%28-1.14%20%5Chat%20i%20%2B%200.33%20%5Chat%20j%29%20%3D%20%28%5Cfrac%7B0.560%7D%7B9.81%7D%29%28v_f%20-%20%2820%5Chat%20i%20-%204%5Chat%20j%29%29)
![-19.97 \hat i + 5.79\hat i + 20 \hat i - 4\hat j = v_f](https://tex.z-dn.net/?f=-19.97%20%5Chat%20i%20%2B%205.79%5Chat%20i%20%2B%2020%20%5Chat%20i%20-%204%5Chat%20j%20%3D%20v_f)
![v_f = 0.03 \hat i + 1.79 \hat j](https://tex.z-dn.net/?f=v_f%20%3D%200.03%20%5Chat%20i%20%2B%201.79%20%5Chat%20j)
Answer: repeatable
Explanation: Accurate repeatable is needed in a valid experiment. I think