Answer:
The x component of the resultant force is -7.27N.
Explanation:
To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:

Since both vectors are in the left side of the y-axis, they have a negative x component. So:

Finally, we sum both components to obtain the component of the resultant force:

In words, the x component of the resultant force is -7.27N.
(a) 328.6 kg m/s
The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

where
m = 62.0 kg is the mass of the passenger
is the change in velocity of the car (and the passenger), which is

So, the linear impulse experienced by the passenger is

(b) 404.7 N
The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

where in this case
is the linear impulse
is the time during which the force is applied
Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m