Cheese I'm so sorry for not having an answer I failed you
Answer:
A=50mΩ
B≅50mΩ
Explanation:
A) To answer this question we have to use the Current Divider Rule. that rule says:
(1)
Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.
We now have a set of two resistor in parallel, so:
(2)
where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).
substituting and rearranging (2)
(3)
Now substituting (3) in (1).

solving this, The value of R1 is: 50mΩ
This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.
Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ
If you substitude this values on (2) Req. will be 49.97 mΩ.
Answer:
50N increasing
Explanation:
it looks like we want the net force. this is the sum of all forces acting on the object. here the problem is extremely simplified and we are even given the values of force. this looks like the rocket is producing an upward force of 160 newtons. it also looks like maybe weight (gravity) and drag (air resistance) are producing downward force against the rocket. we can say those would be subtracting from the amount of upwards force. so 160-40-70=50N. since the force is positive that means we will still have a magnitude of force vector that points upward and so we will increase.
Answer:
t = 0.55[sg]; v = 0.9[m/s]
Explanation:
In order to solve this problem we must establish the initial conditions with which we can work.
y = initial elevation = - 1.5 [m]
x = landing distance = 0.5 [m]
We set "y" with a negative value, as this height is below the table level.
in the following equation (vy)o is equal to zero because there is no velocity in the y component.
therefore:
![y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7Bo%7D%2At%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%5C%5C%20%20%20where%3A%5C%5C%28v_%7By%7D%29_%7Bo%7D%3D0%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%5Bsg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%5D%5C%5C%20-1.5%20%3D%200%2At%20-4.905%2At%5E%7B2%7D%20%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B1.5%7D%7B4.905%7D%20%7D%20%5C%5Ct%3D0.55%5Bs%5D)
Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.
![(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7B0.5%7D%7B0.55%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D0.9%5Bm%2Fs%5D)
106.68 centimetres are in 3.50 feet