From Newton’s Second Law of Motion, the force needed to accelerate a moving object of mass m at acceleration a is given by ;
F = ma
In this case, m = 5kg, a = 6m/s^2 and we want to find F
F = 5 x 6
F = 30N
Therefore, a force of 30N is needed to accelerate a 5kg object at 6m/s^2.
Answer:
Option C
Step-by-step explanation:
I took this before
The question is find the height of the tree, given that at two points 65 feet apart on the same side of the tree and in line with it, the angles of elevaton of the top of the tree are 21° 19' and 16°20'.
1) Convert the angles to decimal form:
19' * 1°/60' = 0.32° => 21° 19' = 21.32°
20' * 1°/60' = 0.33° => 16° 20' = 16.33°
2) Deduce the trigonometric ratios from the verbal information.
You can form a triangle with
- horizontal leg x + 65 feet
- elevation angle 16.33°
- vertical leg height of the tree, h
=> trigonometric ratio: tan (16.33) = h /( x + 65) => h = (x+65) * tan(16.33)
You can form a second triangle with:
- horizontal leg x
- elevation angle 21.32°
- vertical leg height of the tree, h
=> trigonometric ratio: tan(21.32) = h / x => h = x * tan(21.32)
Now equal the two expressions for h:
(x+65)*tan(16.33) = x*tan(21.32)
=> x*tan(16.33) + 65*tan(16.33) = x*tan(21.32)
=> x*tan(21.32) - x*tan(16.33) = 65*tan(16.33)
=> x = 65*tan(16.33) / [ tan(21.32) - tan(16.33) ] = 195.73 feet
=> h = 195.73 * tan(21.32) = 76.39 feet.
Answer: 76.39 feet
H is equal to 3. need the solution?
