<u>Answer:</u>
The force F applied to the handle = 330.03 N
<u>Explanation:</u>
The force can be resolved in to two, horizontal component and vertical component. If θ is the angle between horizontal and applied force we have
Horizontal component of force = F cos θ
Vertical component of force = F sin θ
In this problem normal force exerted on the suitcase is 160 N, that is vertical component of force = 160 N and angle θ = 29⁰.
So, F sin 29 = 160
F = 330.03 N
The force F applied to the handle = 330.03 N
B is the correct answer according to my calculations.
At Z ... slowest speed
At Y ... fastest speed
At X ... medium speed
Wherever it is in its orbit, the line from the planet to the Sun smears over the same amount of area every second.
That's Kepler's second law of planetary motion.
The reason this happens is: That's how gravity works. (A better explanation is available, but first you have to be able to twirl calculus and solid geometry in the air on long sticks.)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

and since the electric field E in between parallel plates separated a distance d and under a potential difference
, is given by:

then :

We want to find when the particle reaches velocity zero via kinematics:

We replace this time (t) in the kinematic equation for the particle displacement:

Replacing the values with the information given, converting the distance d into meters (0.01 m), using
, and the electron's kinetic energy:

we get:
Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
Answer:
Explanation:
For an electric force, F the formula:
F = kQq/r^2
Given:
r2 = 1/2 × r1
F1 × r1 = k
F1 × r1 = F2 × r2
F2 = (F1 × r1^2)/(0.5 × r1)^2
= (F1 × r1^2)/0.25r1^2
= 4 × F1.