What do you mean? I'm confused... You need to put the rest of the question
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
Answer:
The mass of the other worker is 45 kg
Explanation:
The given parameters are;
The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker
The mass of the lighter construction worker, m₁ = 90 kg
The height level of the lighter construction worker's location = h₁
The height level of the other construction worker's location = h₂ = 2·h₁
The gravitational potential energy, P.E., is given as follows;
P.E. = m·g·h
Where;
m = The mass of the object at height
g = The acceleration due to gravity
h = The height at which is located
Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker
We have;
P.E.₁ = P.E.₂
Therefore;
m₁·g·h₁ = m₂·g·h₂
h₂ = 2·h₁
We have;
m₁·g·h₁ = m₂·g·2·h₁
m₁ = 2·m₂
90 kg = 2 × m₂
m₂ = (90 kg)/2 = 45 kg
The mass of the other construction worker is 45 kg.
Answer:
D. Glacial striations
Explanation:
Glacial striations are marks and grooves carved on a rock surface as the materials carried by it abraded the surfaces of the earth.