Question four bulbs A,B,C and D are connected in a circuit shown in the figure below, the letters X, Y and Z represent three switches. Which switch is used to operate switch A separately?
Answer: x
Answer:
1)a. It is constant the whole time the ball is in free-fall.
2)b. = 14 m/s
3) e. = 19.6 m/s
Explanation:
1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.
2) speed = distance/time
horizontal distance = 56m
Time = 4 seconds
Speed = 56m/4s = 14m/s
3) acceleration due to gravity g = 9.8m/s^2
Initial vertical velocity = u
Final vertical velocity = v = -u
Using the law of motion;
v = u + at
a = acceleration = -g = -9.8m/s^2
t = time of flight = 4
Substituting the values;
-u = u - 4(9.8)
-2u = -4(9.8)
u = -4(9.8)/-2
u = 2(9.8) = 19.6 m/s
Initial vertical velocity = u = 19.6 m/s
I believe that it is the first one just a guess tho. So don't trust me, just in case
Let's use Newton's 2nd law of motion:
Force = (mass) x (acceleration)
Force = (68 kg) x (1.2 m/s²) = 81.6 newtons .
As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that
![F = \frac{kq_1q_3}{d_{13}^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bkq_1q_3%7D%7Bd_%7B13%7D%5E2%7D)
now from the above equation
![F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%289%5Ctimes%2010%5E9%29%282%5Ctimes%2010%5E%7B-6%7D%29%284%20%5Ctimes%2010%5E%7B-6%7D%29%7D%7B0.5%5E2%7D)
![F = 0.288 N](https://tex.z-dn.net/?f=F%20%3D%200.288%20N)
so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by
![F_{net} = 2Fcos\theta](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%202Fcos%5Ctheta)
here we know that angle is
![\theta = 37 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2037%20degree)
now we have
![F_{net} = 2\times 0.288 cos37](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%202%5Ctimes%200.288%20cos37)
![F_{net} = 0.46 N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%200.46%20N)
so net force on q3 is 0.46 N vertically upwards along +Y axis