Answer:
the horizontal distance is 4.355 meters
Explanation:
The computation of the horizontal distance while travelling in the air is shown below:
Data provided in the question is as follows
Velocity = u = 7.70 m/s
H = 1.60 m
R = horizontal direction
Based on the above information
As we know that
R = u × time
where,
Time =
So,
=
= 4.355 meters
hence, the horizontal distance is 4.355 meters
Complete Question
The complete question shown on the first uploaded image
Answer:
a)
The force on Q due to dipole is Attractive
b)
The charge Q exerts attractive force on the dipole
c)
Yes from the above parts, force depends on the sign of charge
d)
e)
The magnitude o force decrease by a factor of 8.0 times
Explanation:
The explanation is shown on the second uploaded image
Answer:
(a) 196 m/s
(b) 490 m
(c) 3394.82 m
(d) 2572.5 m
Explanation:
First of all, let us know one thing. When an object is thrown in the air, it experiences two forces acting in two different directions, one in the horizontal direction called air resistance and the second in the vertically downward direction due to its weight. In most of the cases, while solving numerical problems, air resistance is neglected unless stated in the numerical problem. This means we can assume zero acceleration along the horizontal direction.
Now, while solving our numerical problem, we will discuss motion along two axes according to our convenience in the course of solving this problem.
<u>Given:</u>
<u>Assume:</u>
Let us assume that the position from where the projectile was projected lies at origin.
Part (a):
During the time of flight the displacement of the projectile along the vertical is zero as it comes to the same vertical height from where it was projected.
Hence, the initial speed of the projectile is 196 m/s.
Part (b):
For a projectile, the time take by it to reach its maximum height is equal to return from the maximum height to its initial height is the same.
So, time taken to reach its maximum height will be equal to 10 s.
And during the upward motion of this time interval, the distance travel along the vertical will give us maximum height.
Hence, the maximum altitude is 490 m.
Part (c):
Range is the horizontal displacement of the projectile from the initial position. As acceleration is zero along the horizontal, the projectile is in uniform motion along the horizontal direction.
So, the range is given by:
Hence, the range of the projectile is 3394.82 m.
Part (d):
In order to calculate the displacement of the projectile from its initial position, we first will have to find out the height of the projectile and its range during 15 s.
Hence, the displacement from the point of launch to the position on its trajectory at 15 s is 2572.5 m.