The higher the pressure, the higher boiling point of water. At lower the pressure, the boiling point of water comes down. So, the lower pressure inreases the boiling resulting more evaporation. As we go higher in altitude, the atmospheric pressure decreases. This results in decreasing the boiling point at higher altitude and increase in boiling of water. In fact, at the sea level ,the the sea water boils at 100 degree C where atmospheric pressre is normal. However , the boiling takes place at a lower temperature at the top of a mountain due to low pressure. In other words the boling is faster at the top of a mountain than that at its foot.
Answer: the rarest element is Francium. J is not on the periodic table. also Dmitri Mendeleev proposed the periodic table.
Explanation: Kinda looked the last one up.
Explanation:
energy conservation and fatigue management -tiredness is a common symptom of a heart attack and although rest is important activity is also required to facilitate a return to health. an occupational therapist said energy conservation and fatigue management is techniques to be implemented throughout the day. to help clients achieve their goals
Answer:
v = 15.8 m/s
Explanation:
Let's analyze the situation a little, we have a compressed spring so it has an elastic energy that will become part kinetic energy and a potential part for the man to get out of the barrel, in addition there is a friction force that they perform work against the movement. So the variation of mechanical energy is equal to the work of the fictional force
= ΔEm = 
-Em₀
Let's write the mechanical energy at each point
Initial
Em₀ = Ke = ½ k x²
Final
= K + U = ½ m v² + mg y
Let's use Hooke's law to find compression
F = - k x
x = -F / k
x = 4400/1100
x = - 4 m
Let's write the energy equation
fr d = ½ m v² + mgy - ½ k x²
Let's clear the speed
v² = (fr d + ½ kx² - mg y) 2 / m
v² = (40 4.00 + ½ 1100 4² - 60.0 9.8 2.50) 2/60.0
v² = (160 + 8800 - 1470) / 30
v = √ (229.66)
v = 15.8 m/s
Answer:
a. v₁ = 16.2 m/s
b. μ = 0.251
Explanation:
Given:
θ = 15 ° , r = 100 m , v₂ = 15.0 km / h
a.
To determine v₁ to take a 100 m radius curve banked at 15 °
tan θ = v₁² / r * g
v₁ = √ r * g * tan θ
v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s
b.
To determine μ friction needed for a frightened
v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg
v₂ = 4.2 m/s
fk = μ * m * g
a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²
a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²
F₁ = m * a₁ , F₂ = m * a₂
fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)
μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g
μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)
μ = 0.251
