Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.
From Newton's second law we understand that
Gravity at this case)
Where,
m = mass
a= acceleration
Also we know that

Part A) The buoyant force acting on the balloon is given as

As mass is equal to the density and Volume and acceleration equal to Gravity constant



PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then




PART C) The additional mass that can the balloon support in equilibrium is given as




Answer:It is an Endothermic reaction
Explanation:
Exothermic reaction
An exothermic reaction is one that releases energy in the form of heat or light.
Endothermic reaction
An endothermic reaction is one that absorbs energy in the form of heat or light.
Answer: option "A" is correct
Explanation: the closet to understand the intensity level will be from 80db (decibels) upwards. So option a which is 84db is the answer. Thank you.
Answer:
b. Static > sliding > rolling friction.
Explanation:
Static friction is greater than sliding friction. It takes more force to get an object to start sliding than to keep it sliding.
Sliding friction is greater than rolling friction. There are fewer points of contact for a round surface compared to a flat one.