Answer: A
Explanation:
honestly, it sounded the best
Answer:
Wn = 9.14 x 10¹⁷ N
Explanation:
First we need to find our mass. For this purpose we use the following formula:
W = mg
m = W/g
where,
W = Weight = 675 N
g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²
m = Mass = ?
Therefore,
m = (675 N)/(9.8 m/s²)
m = 68.88 kg
Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:
gn = (G)(Mn)/(Rn)²
where,
gn = acceleration due to gravity on surface of neutron star = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg
Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m
Therefore,
gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)
gn = 13.27 x 10¹⁵ m/s²
Now, my weight on neutron star will be:
Wn = m(gn)
Wn = (68.88)(13.27 x 10¹⁵ m/s²)
<u>Wn = 9.14 x 10¹⁷ N</u>
If the machine's mechanical advantage is 4.5, that means that
Output force = (4.5) x (Input force) .
We know the input force, and we need to find the output force. Rather than wander around the room looking at the floor while our hair smolders, let's try putting the numbers we know into the equation I wrote up there. OK ?
Output force = (4.5) x (Input force)
Output force = (4.5) x (800 N)
Now dooda multiplication:
<em>Output force = 3,600 N</em> .
That's exactly what the question asked for. So we're done !
To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:
Where,
l = length of the thread
= circumference of spool
For \phi we have that,
For l we have that
l = 62.8m
Finally the number of Turns would be,
Therefore the number of turns of thread on the spool are 1000turns.
