Answer:
between 10 and 15 percent
Explanation:
How to put your load
- First load the heavy
The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.
In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.
- Hold your load
After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.
- Trailer weight
To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.
GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.
That's the 'electrostatic' force.
Answer:
The horizontal component of the velocity is 188 m/s
The vertical component of the velocity is 50 m/s.
Explanation:
Hi there!
Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:
We can find vx using the following trigonometric rule of a right triangle:
cos α = adjacent / hypotenuse
cos 15° = vx / 195 m/s
195 m/s · cos 15° = vx
vx = 188 m/s
The horizontal component of the velocity is 188 m/s
To calculate the y-component we will use the following trigonometric rule:
sin α = opposite / hypotenuse
sin 15° = vy / 195 m/s
195 m/s · sin 15° = vy
vy = 50 m/s
The vertical component of the velocity is 50 m/s.
The point of contact the path difference is zero but one of the interfering ray is reflected so the effective path difference becomes λ/2 thus the condition of minimum intensity is created in the center.
