Answer:
According to the Newton's law of gravitational every object in the universe attracts every other objects with a force which is called gravitational force.This gravitational force is (i) directly proportional to the product of their masses and (ii) inversely proportional to the square of the distance between their centres.
Explanation:
Newton's law of gravitational is called the universal law because it is applicable to all the bodies either terrestrial or celestial having any shape,size,mass or at any distance apart with any medium between them,at any time(past,present or future).
 
        
             
        
        
        
Answer:
Thus, the change in the weight of the person is 1.6N , option c is correct.
Explanation:
 
        
             
        
        
        
<h2>
Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.</h2>
Explanation:
Let speed of motor boat be m and speed of current be c.
A motorboat traveling with a current can go 160 km in 4 hours.
    Distance = 160 km
    Time = 4 hours
     Speed = m + c
    We have
             Distance = Speed x Time
             160 = (m+c) x 4
             m + c = 40     --------------------- eqn 1
Against the current it takes 5 hours to go the same distance.
    Distance = 160 km
    Time = 5 hours
     Speed = m - c
    We have
             Distance = Speed x Time
             160 = (m-c) x 5
             m - c = 32     --------------------- eqn 2
eqn 1 + eqn 2
            2m = 40 + 32
              m = 36 km/hr
Substituting in eqn 1
                36 + c = 40
                       c = 4 km/hr
Speed of motorboat is 36 km/hr and speed of current is 4 km/hr.
 
        
        
        
For number 11, you should say that there is more pollution and fossil fuels being burned.
        
             
        
        
        
Answer:
0.546 
Explanation:
From the given information:
The force on a given current-carrying conductor is:

where the length usually in negative (x) direction can be computed as 

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:



![F = I  (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%5E3_1%20%5Chat%20k)
![F = I  (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ]  \hat k](https://tex.z-dn.net/?f=F%20%3D%20I%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B3%5E3%7D%7B3%7D%20-%20%5Cdfrac%7B1%5E3%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
where;
current I = 7.0 A
![F = (7.0 \ A)  (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ]  \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B27%7D%7B3%7D%20-%20%5Cdfrac%7B1%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
![F = (7.0 \ A)  (9.0) \bigg [\dfrac{26}{3} \bigg ]  \hat k](https://tex.z-dn.net/?f=F%20%3D%20%287.0%20%5C%20A%29%20%20%289.0%29%20%5Cbigg%20%5B%5Cdfrac%7B26%7D%7B3%7D%20%5Cbigg%20%5D%20%20%5Chat%20k)
F = 546 × 10⁻³ T/mT 
F = 0.546 