Answer:
F'= 4F/9
Explanation:
Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :
...(1)
Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :
.....(2)
Dividing equation (1) and (2), we get :
Hence, the correct option is (d) i.e. " 4F/9"
Answer:
1. v = 6.67 m/s
2. d = 9.54 m
Explanation:
1. To find the horizontal velocity of the rock we need to use the following equation:
<u>Where</u>:
d: is the distance traveled by the rock
t: is the time
The time can be calculated as follows:
<u>Where:</u>
g: is gravity = 9.8 m/s²
Now, the horizontal velocity of the rock is:
Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.
2. To calculate the distance at which the projectile will land, first, we need to find the time:
So, the distance is:
Therefore, the projectile will land at 9.54 m of the second cliff.
I hope it helps you!
Answer:
82.1 km
Explanation:
We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:
Vector B is 48 km south, so:
Finally, vector C:
Now we add the components along each direction:
So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is