Answer:
Acceleration of the bullet will be 1778835.6
Explanation:
We have given length of the barrel refile s= 0.855 m
When the bullet leaves the muzzle its velocity is 553 m/sec
So final velocity v = 553 m/sec
Initial velocity will be 0 that is u = 0 m/sec
According to third equation of motion
The package should be dropped <u>678 m</u> short of the target.
A package dropped from a plane which is moving at a speed v, has a horizontal velocity equal to the horizontal velocity of the plane. It has a parabolic trajectory, traversing a horizontal range x while it falls through a vertical height y.
The package has no initial vertical velocity, and it falls through a height y under the action of the acceleration due to gravity g.
Use the equation,
Write an expression for t, the time taken for the package to fall through y.
Substitute 100 m for y and 9.81m/s² for g.
In the time t the package travels a horizontal distance x. The horizontal velocity of the package remains constant, since no force acts along the horizontal direction.
Therefore, the horizontal distance traveled by the package is given by,
If the package is released <u>678m</u> before the target, the package would reach the scientists working in Greenland.
Answer:
The force felt by charge 3 is F=(-5.6*10⁻⁶,3.36⁻⁵)N
Explanation:
As the superposition principle applies to static charges, we can find the net electric force as the sum of the two forces felt by q3.
Looking at the drawing and knowing that they form an equilateral triangle of lenght 4 we can conclude that each internal angle is 60°.
So, the positions in our coordinate system are:
Now using Coulomb's force:
Where d=4, q1 = -7.8*10⁻⁹C, q2 = -15.6 *10⁻⁹C, q3 = 8.0 *10⁻⁹C, k=8.98*10⁹, e0=8.8*10¹⁰:
Replacing we get 2 equations:
To work with the sam
F=∑F_i=3.5*10⁻⁴(0.023,0.032)+7*10⁻⁴(-0.016,0.032)=
=((3.5*10⁻⁴-7*10⁻⁴)*0.016,(3.5*10⁻⁴+7*10⁻⁴)*0.032)=
F=(-5.6*10⁻⁶,3.36⁻⁵)N
