The distance of separation between the two masses is 0.927 m.
<h3>Gravitational force:</h3>
This is the force that exists between two masses in the universe.
To calculate the distance of separation of the masses, we use the formula below.
- F = GMm/r².............. Equation 1
Where:
- F = Gravitational force
- m = First mass
- M = Second mass
- G = Universal constant
- r = distance of seperation.
Make r the subject of the equation.
- r = √(GMm/F)................... Equation 2
From the question,
Given:
- F = 3.3×10⁻⁷ N
- m = 61 kg
- M = 75 kg
- G = 6.69×10⁻¹¹ Nm²/kg²
Substitute these values into equation 2
- r = √(61×75×6.69×10⁻¹¹)/(3.3×10⁻⁷)
- r = 0.927 m
Hence, The distance of separation between the two masses is 0.927 m
Learn more about Gravitational force here: brainly.com/question/11359658
Their chemicals and their reactions, hope I helped
The terminal velocity as it falls through still air is 4.65154 in/s.
The diameter of small water droplet is 1.25 mil= 1.25×0.0254×10^-3 m
= 3.175 × 10^-5 m
Now the viscosity of still air is η = 1.83× 10⁻⁵ Pa
So the formula for drag force is:
Fd = 6πηrv
where, v is the velocity.
Now to attain terminal velocity acceleration must be zero.
→ W = Fd
ρVg = 6πrηv
ρ × 4/3 πr³×g = 6πrηv
v = 2/9 × ρgr³/ η
v = 2/9 × 10³×9.81×(3.175×10^-3) / 18.6×10^-6
v = 0.1181 m/s
v = 4.65154 in/s
Learn more about terminal velocity here:
brainly.com/question/20409472
#SPJ4
<h2>
Answer:5
,133.6
,51.18
</h2>
Explanation:
Let 
,
be the horizontal and vertical components of velocity.
Question a:
Horizontal component of velocity is the ratio of range and time of flight.
So,horizontal component of velocity is 
So,
Question b:
Time of flight=
So,
Maximum height is given by 
So,maximum height is 
Question c:
The vertical velocity is already calculated in Question b.
Sorry don’t know this one
