Answer:b.20.8L
Explanation:
Since pressure is constant we can use the equation V1/T1 = V2/T2
(20/250)=(x/260)
5200=250x
X=20.8
Layer F was formed earlier than Layer A
Answer:
The acceleration of the player is - 4.9 m/s²
Explanation:
The given is:
1. The mass of the player is 55 kg
2. His initial speed is 4.6 m/s
3. The coefficient of the kinetic fraction between the player and the
ground is 0.50
We need to find the player acceleration
According to Newton's Law
→ ∑ forces in direction of motion = mass × acceleration
There is only the friction force opposite to the motion
→ Friction force = μR
where μ is the coefficient of friction and R is the normal reaction
→ The normal reaction R = mg
where m is the mass and g is the acceleration of gravity
→ m = 55 kg , g = 9.8 m/s²
→ R = 55 × 9.8 = 539 N
→ ∑ F = - μR
→ - μR = m × a
→ μ = 0.5 , R = 539 N , m = 55
→ -(0.5)(539) = 55 × a
→ - 269.5 = 55 a
Divide both sides by 55
→ a = - 4.9 m/s²
The acceleration of the player is - 4.9 m/s²
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Answer:
the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers
Explanation:
The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

15 - (1/200) L = 5 + (1/200) L
15 - 5 = (1/200) L + (1/200) L
10 = (2/200) L
(10*200)/2 = L
1000 = L
Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:
W = 15 - (1/200) L
W = 15 - (1/200) 1000
W = 10
Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers