Answer:
191.36 N/m
Explanation:
From the question,
The Potential Energy of the safe = Energy of the spring when it was compressed.
mgh = 1/2ke²............... Equation 1
Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression
Making k the subject of the equation,
k =2mgh/e²................ Equation 2
Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m
Constant: g = 9.8 m/s²
Substitute into equation 2
k = 2(1100)(9.8)(0.0024)/0.52²
k = 51.744/0.2704
k = 191.36 N/m
Hence the spring constant of the heavy-duty spring = 191.36 N/m
Answer:
105 mg
Explanation:
Given that:
1 baked potato provides 30 mg of vitamin C.
So,
70 baked potatoes provide 
mg of vitamin C
Also,
70 potatoes = 20 lb
So,
20 lb potatoes provide mg of vitamin C
Thus,
1 lb potatoes provide 
mg of vitamin C
<u>Thus, 105 mg of Vitamin C are provided per pound of the potatoes.</u>
Answer:
As the launch force increase the launch velocity will
<em><u>Increase</u></em>
The reason for your answer to number six is because
<em><u>There is a direct relationship between force and acceleration.</u></em>
<em><u /></em>
Explanation:
<em>It is known all over the place that, there is a direct relationship between Force and acceleration of an object leading to an increase in force being directly proportional to the increase in the acceleration of the given object and vice versa.</em>
<span>anwser will be
F = ma
where
F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet
Substituting appropriately,
F = 0.005a --- call this Equation 1
Next working equation is
Vf^2 - Vo^2 = 2as
where
Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel
Substituting appropriately,
326^2 - 0 = 2(a)(0.83)
a = 64,022 m/sec^2
the anwser will be
Substituting this into Equation 1,
F = 0.005(64,022)
F =320.11 Newtons
Hope this helps. </span><span>
</span>
