Answer: The forces are balanced.
Answer:
A. The object falls a distance of 250 m
Explanation:
Hi there!
In the question, you have forgotten the acceleration due to gravity. However, looking on the web I´ve found a very similar problem in which the acceleration due to gravity was as twice as much as it is on Earth.
The equation of height of a falling object is the following:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the object after a time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (on Earth: ≅ -10 m/s² considering the upward direction as positive).
Let´s place the origin of the system of reference at the point where the object is released so that y0 = 0. Since the object falls from rest, v0 = 0.
Then, the height of the object after 5 s will be :
y = 1/2 · 2 · g · t² (notice that the acceleration due to gravity is 2 · g)
y = g · t²
y = -10 m/s² · (5 s)²
y = -250 m
The object falls a distance of 250 m.
Answer:
h = 618.64 m
Explanation:
First we need to calculate the height gained by rocket while the fuel is burning. We use 2nd equation of motion for that purpose:
h₁ = Vit + (1/2)at²
where,
h₁ = height gained during the burning of fuel
Vi = Initial Velocity = 0 m/s
t = time = 7 s
a = acceleration = 8 m/s²
Therefore,
h₁ = (0 m/s)(7 s) + (1/2)(8 m/s²)(7 s)²
h₁ = 196 m
Now we use 1st equation of motion to find final speed Vf:
Vf = Vi + at
Vf = 0 m/s + (8 m/s²)(7 s)
Vf = 56 m/s
Now, we calculate height covered in free fall motion. Using 3rd equation of motion:
2ah₂ = Vf² - Vi²
where,
a = - 3.71 m/s²
h₂ = height gained during free fall motion = ?
Vf = Final Velocity = 0 m/s (since, rocket will stop at highest point)
Vi = 56 m/s
Therefore,
(2)(-3.71 m/s²)h₂ = (0 m/s)² - (56 m/s)²
h₂ = 422.64 m
So the total height gained will be:
h = h₁ + h₂
h = 196 m + 422.64 m
<u>h = 618.64 m</u>
Answer:
v₂ = 79.69 m/s
Explanation:
The initial diameter of the hose, d₁ = 3.0 cm = 0.03 m
Initial Cross Sectional Area, A₁ = πd₁²/4
A₁ = (π* 0.03²)/4
A₁ = 0.00071 m²
The initial speed of water from the hose, v₁ = 2.2 m/s
The diameter of the hose after blocking the end, d₂ = 0.50 cm = 0.005 m
Cross Sectional Area of the hose after blocking the end, A₂ = πd₂²/4
A₂ = (π* 0.005²)/4
A₂ = 0.0000196 m²
To get the speed, v₂, at which the water spray from the hose after blocking the end, we will use the continuity equation:
A₁v₁ = A₂v₂
0.00071 * 2.2 = 0.0000196 v₂
0.001562 = 0.0000196 v₂
v₂ = 0.001562/0.0000196
v₂ = 79.69 m/s
That would be the mass number for that certain isotope.
Also is the <span>sum of protons and neutrons in the atomic nuclei of the isotope.</span>