Answer:
60 N
Explanation:
because when we double the 30N, we will get 60N as a force
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is
Explanation:
From the question we are told that
The third charge is ![q_3 = 55 nC = 55 *10^{-9} C](https://tex.z-dn.net/?f=q_3%20%3D%20%2055%20nC%20%3D%20%2055%20%2A10%5E%7B-9%7D%20C)
The position of the third charge is ![x = -1.220 \ m](https://tex.z-dn.net/?f=x%20%3D%20-1.220%20%5C%20m)
The first charge is ![q_1 = -16 nC = -16 *10^{-9} \ C](https://tex.z-dn.net/?f=q_1%20%3D%20%20-16%20nC%20%20%3D%20%20-16%20%2A10%5E%7B-9%7D%20%5C%20C)
The position of the first charge is ![x_1 = -1.650m](https://tex.z-dn.net/?f=x_1%20%3D%20%20-1.650m)
The second charge is ![q_2 = 32 nC = 32 *10^{-9} C](https://tex.z-dn.net/?f=q_2%20%3D%20%2032%20nC%20%20%3D%20%2032%20%2A10%5E%7B-9%7D%20C)
The position of the second charge is
The distance between the first and the third charge is
![d_{1-3} = -1.650 -(-1.220)](https://tex.z-dn.net/?f=d_%7B1-3%7D%20%3D%20%20-1.650%20-%28-1.220%29)
![d_{1-3} = -0.43 \ m](https://tex.z-dn.net/?f=d_%7B1-3%7D%20%3D%20-0.43%20%5C%20m)
The force exerted on the third charge by the first is
![F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}](https://tex.z-dn.net/?f=F_%7B1-3%7D%20%3D%20%20%5Cfrac%7Bk%20%20q_1%20q_3%7D%7Bd_%7B1-3%7D%5E2%7D)
Where k is the coulomb's constant with a value ![9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.](https://tex.z-dn.net/?f=9%2A10%5E%7B9%7D%20%5C%20kg%5Ccdot%20m%5E3%5Ccdot%20s%5E%7B-4%7D%5Ccdot%20A%5E2.)
substituting values
The distance between the second and the third charge is
![d_{2-3} = 0- (-1.22)](https://tex.z-dn.net/?f=d_%7B2-3%7D%20%3D%20%200-%20%28-1.22%29)
![d_{2-3} =1.220 \ m](https://tex.z-dn.net/?f=d_%7B2-3%7D%20%3D1.220%20%5C%20m)
The force exerted on the third charge by the first is mathematically evaluated as
substituting values
![F_{2-3} = 1.06*10^{-5} N](https://tex.z-dn.net/?f=F_%7B2-3%7D%20%3D%20%201.06%2A10%5E%7B-5%7D%20N)
The net force is
substituting values
![F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%204.28%20%2A10%5E%7B-5%7D%20-%201.06%2A10%5E%7B-5%7D)
Complete Question:
Check the file attached to get the complete question
Answer:
In the film Ice word Revenge, vehicle 2 did not fall of the cliff because,
but in Claire's test, vehicle 2 off the cliff because ![Weight_{vehicle 1} \geq Weight_{vehicle 2}](https://tex.z-dn.net/?f=Weight_%7Bvehicle%201%7D%20%5Cgeq%20Weight_%7Bvehicle%202%7D)
Explanation:
In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff. In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.
1250 J in 5 sec= 250 Joule(s) per second (1250/5 0
250 Joules per second = 250 Watts ( 1J/s = 1 Watt per definition)
250 Watts output = 250/0.65 efficiency = 384 Watts input
1 Horsepower = 732 Watts
Motors 1 Horsepower and under are made in certain step sizes like
3/4 , 1/2 , 1/3, 1/4, 1/16 1/20 of a Horsepower.
3/4 Horsepower is 549 Watts
1/2 Horsepower is 366 Watts
so you need to 3/4 horsepower motor to achieve 1250 J of work in 5 seconds.
Answer:
<h2>480</h2>
Explanation:
<h2>R=120÷0.25</h2><h2>R=480 ohms </h2>
because the unit for resistance is in ohms