Answer:
13.33 or 13 1/3m/s (meters per second)
Explanation:
In physics, we use the basic units of meters and seconds. So first convert (km) into meters (m) and also hours and minutes into seconds (s). We end up with 120000m and 9000s. Then divide the 120000m by the 9000s and you end up with 13.33 or 13 1/3 m/s.
Answer:
1)
is<u> positive.</u>
<u></u>
2) ![\rm q_2=4.56\times 10^{-10}\ C.](https://tex.z-dn.net/?f=%5Crm%20q_2%3D4.56%5Ctimes%2010%5E%7B-10%7D%5C%20C.)
Explanation:
<h2><u>
Part 1:</u></h2>
<u></u>
The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.
Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e.,
is <u>positive.</u>
<u></u>
<h2><u>
Part 2:</u></h2>
<u></u>
<u>Given:</u>
- Mass of the balloon, m = 0.00275 kg.
- Charge on the balloon,
![\rm q_1 = -3.50\times 10^{-8}\ C.](https://tex.z-dn.net/?f=%5Crm%20q_1%20%3D%20-3.50%5Ctimes%2010%5E%7B-8%7D%5C%20C.)
- Distance between the rod and the balloon, d = 0.0640 m.
- Acceleration due to gravity,
![\rm g = 9.81\ m/s^2.](https://tex.z-dn.net/?f=%5Crm%20g%20%3D%209.81%5C%20m%2Fs%5E2.)
In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.
Weight of the balloon, ![\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.](https://tex.z-dn.net/?f=%5Crm%20W%20%3D%20mg%20%3D%200.00275%5Ctimes%209.81%3D2.70%5Ctimes%2010%5E%7B-2%7D%5C%20N.)
The magnitude of the electrostatic force on the balloon due to the rod is given by
![\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.](https://tex.z-dn.net/?f=%5Crm%20F_e%20%3D%20%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%5Cdfrac%7B%7Cq_1%7C%7Cq_2%7C%7D%7Bd%5E2%7D.)
is the Coulomb's constant.
For the elecric force and the weight to be balanced,
![\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.](https://tex.z-dn.net/?f=%5Crm%20F_e%20%3D%20W%5C%5C%5Cdfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_o%7D%5Cdfrac%7B%7Cq_1%7C%7Cq_2%7C%7D%7Bd%5E2%7D%3DW%5C%5C8.99%5Ctimes%2010%5E9%5Ctimes%20%5Cdfrac%7B3.50%5Ctimes10%5E%7B-8%7D%5Ctimes%20%7Cq_2%7C%20%7D%7B0.0640%5E2%7D%3D2.70%5Ctimes%2010%5E%7B-2%7D%5C%5C%7Cq_2%7C%20%3D%20%5Cdfrac%7B2.70%5Ctimes%2010%5E%7B-2%7D%5Ctimes%200.00640%5E2%7D%7B8.99%5Ctimes%2010%5E9%5Ctimes%202.70%5Ctimes%2010%5E%7B-7%7D%7D%3D4.56%5Ctimes%2010%5E%7B-10%7D%5C%20C.)
Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here
Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)
Answer:
Resultant Force=33.8 lb
Angle=67.2°
Explanation:
Given data
Fa=22 lb
Fb=16 lb
Θ=55⁰
To find
(i) Resultant Force F
(ii)Angle α
Solution
First we need to represent the forces in vector form
![\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20F_%7B1%7D%3D22j%5C%5C%20F_%7B2%7D%3Du%2Bv%5C%5CF_%7B2%7D%3D16sin%2855%29i%2B16cos%2855%29j%5C%5CF_%7B2%7D%3D16%280.82%29i%2B16%280.5735%29j%5C%5CF_%7B2%7D%3D13.12i%2B9.176j)
Total Force
![F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j](https://tex.z-dn.net/?f=F%3DF_%7B1%7D%2BF_%7B2%7D%5C%5C%20F_%7B2%7D%3D22j%2B13.12i%2B9.176j%5C%5CF_%7B2%7D%3D13.12i%2B31.176j)
The Resultant Force is given as
![|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb](https://tex.z-dn.net/?f=%7CF%7C%3D%5Csqrt%7Bx%5E%7B2%7D%20%2By%5E%7B2%7D%20%7D%5C%5C%7CF%7C%3D%5Csqrt%7B%2813.12%29%5E%7B2%7D%20%2B%2831.176%29%5E%7B2%7D%20%7D%5C%5C%20%7CF%7C%3D33.8lb)
For(ii) angle
We can find the angle bu using tanα=y/x
So
![tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}](https://tex.z-dn.net/?f=tan%5Calpha%20%3D%5Cfrac%7B31.176%7D%7B13.12%7D%5C%5C%20%5Calpha%20%3Dtan%5E%7B-1%7D%20%28%5Cfrac%7B31.176%7D%7B13.12%7D%29%5C%5C%5Calpha%20%3D67.2%5E%7Bo%7D)
Answer:
Resistance = 252.53 Ohms
Explanation:
Given the following data;
Charge = 0.125 C
Voltage = 5 V
Time = 6.3 seconds
To find the resistance;
First of all, we would determine the current flowing through the battery;
Quantity of charge, Q = current * time
0.125 = current * 6.3
Current = 0.125/6.3
Current = 0.0198 A
Next, we find the resistance;
Resistance = voltage/current
Resistance = 5/0.0198
Resistance = 252.53 Ohms
Answer:
weaker has the heavier of an object