Split up each force into horizontal and vertical components.
• 300 N at N30°E :
(300 N) (cos(30°) i + sin(30°) j)
• 400 N at N60°E :
(400 N) (cos(60°) i + sin(60°) j)
• 500 N at N80°E :
(500 N) (cos(80°) i + sin(80°) j)
The resultant force is the sum of these forces,
∑ F = (300 cos(30°) + 400 cos(60°) + 500 cos(80°)) i
… … … + (300 sin(30°) + 400 sin(60°) + 500 sin(80°)) j N
∑ F ≈ (546.632 i + 988.814 j) N
so ∑ F has a magnitude of approximately 1129.85 N and points in the direction of approximately N61.0655°E.
Answer:
22.38 g of silicone-32 will be present in 300 years.
Step-by-step explanation:
A radioactive half-life refers to the amount of time it takes for half of the original isotope to decay and its given by
where,
= quantity of the substance remaining
= initial quantity of the substance
= time elapsed
= half life of the substance
From the information given we know:
- The initial quantity of silicone-32 is 30 g.
- The time elapsed is 300 years.
- The half life of silicone-32 is 710 years.
So, to find the quantity of silicone-32 remaining we apply the above equation
22.38 g of silicone-32 will be present in 300 years.
You can set up an system of equations
x=2y+8
x-y=25
Substitute x in to the second equation
2y+8-y=25
y+8=25
y=17
Substitute the y back in to the first equation.
x=2(17)+8=42
I have no idea what did he do
