Answer: The original temperature was
Explanation:
Let's put the information in mathematical form:
If we consider the helium as an ideal gas, we can use the Ideal Gas Law:
were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)
From this, taking , we have:
⇒
Now:
⇒
gravitational force of planet exerted on its object near the surface is known as weight
so here we know that gravitational force of mars is much less than the gravitational force of Earth
So on the surface of mars the Weight of objects must be much less than the weight of object on surface of Earth
so here correct answer must be
<em>D. Your weight would decrease.</em>
To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:
Where,
P = Axial Load
l = Gage length
A = Cross-sectional Area
E = Modulus of Elasticity
Our values are given as,
l = 3.5m
D = 0.028m
E = 200GPa
Replacing we have,
Therefore the change in length is 1.93mm
Answer:
v = 2.82 m/s
Explanation:
For this exercise we can use the conservation of energy relations.
We place our reference system at the point where block 1 of m₁ = 4 kg
starting point. With the spring compressed
Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂
final point. When block 1 has descended y = - 0.400 m
Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y
as there is no friction, the energy is conserved
Em₀ = Em_f
½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y
½ k x² - m₁ g y = ½ m₂ v²
v² =
let's calculate
v² =
v² = 2.7 + 5.23
v = √7.927
v = 2,815 m / s
using of significant figures
v = 2.82 m/s