Answer:
The value is ![A = 0.014 \ m](https://tex.z-dn.net/?f=A%20%20%3D%20%200.014%20%5C%20%20m)
Explanation:
From the question we are told that
The mass of the object is ![m = 2.0 \ kg](https://tex.z-dn.net/?f=m%20%20%3D%20%202.0%20%5C%20%20kg)
The unstressed length of the string is ![l = 0.08 \ m](https://tex.z-dn.net/?f=l%20%20%3D%20%200.08%20%5C%20%20m)
The length of the spring when it is at equilibrium is ![l_e = 5.9 \ cm = 0.059 \ m](https://tex.z-dn.net/?f=l_e%20%3D%205.9%20%5C%20%20cm%20%20%3D%20%200.059%20%5C%20%20m)
The initial speed (maximum speed)of the spring when given a downward blow ![v = 0.30 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%200.30%20%5C%20%20m%2Fs)
Generally the maximum speed of the spring is mathematically represented as
![u = A * w](https://tex.z-dn.net/?f=u%20%3D%20%20A%20%2A%20%20w)
Here A is maximum height above the floor (i.e the maximum amplitude)
and
is the angular frequency which is mathematically represented as
![w = \sqrt{\frac{k}{m} }](https://tex.z-dn.net/?f=w%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D)
So
![u = A * \sqrt{\frac{k}{m} }](https://tex.z-dn.net/?f=u%20%3D%20%20A%20%2A%20%20%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%20%7D)
=> ![A = u * \sqrt{\frac{m}{k} }](https://tex.z-dn.net/?f=A%20%20%3D%20%20u%20%2A%20%20%20%5Csqrt%7B%5Cfrac%7Bm%7D%7Bk%7D%20%7D)
Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as
![b = l -l_e](https://tex.z-dn.net/?f=b%20%20%3D%20%20l%20-l_e)
=> ![b = 0.08 - 0.05 9](https://tex.z-dn.net/?f=b%20%20%3D%200.08%20-%200.05%209)
=> ![b = 0.021 \ m](https://tex.z-dn.net/?f=b%20%20%3D%200.021%20%5C%20%20m)
Generally at equilibrium position the net force acting on the spring is
![k * b - mg = 0](https://tex.z-dn.net/?f=k%20%2A%20%20b%20%20-%20%20mg%20%20%3D%20%200)
=> ![k * 0.021 - 2 * 9.8 = 0](https://tex.z-dn.net/?f=k%20%2A%20%200.021%20%20%20-%20%20%202%20%2A%209.8%20%20%3D%20%200)
=> ![k = 933 \ N/m](https://tex.z-dn.net/?f=k%20%3D%20%20933%20%5C%20%20N%2Fm)
So
![A = 0.30 * \sqrt{\frac{2}{933} }](https://tex.z-dn.net/?f=A%20%20%3D%20%200.30%20%20%2A%20%20%20%5Csqrt%7B%5Cfrac%7B2%7D%7B933%7D%20%7D)
=> ![A = 0.014 \ m](https://tex.z-dn.net/?f=A%20%20%3D%20%200.014%20%5C%20%20m)
Answer:
<h2>39.2 m</h2>
Explanation:
The height of the hill side can be found by using the formula
![h = \frac{p}{m} \\](https://tex.z-dn.net/?f=h%20%3D%20%20%5Cfrac%7Bp%7D%7Bm%7D%20%20%5C%5C%20)
p is the potential energy
m is the mass
From the question we have
![h = \frac{1568}{40} = \frac{196}{5} \\](https://tex.z-dn.net/?f=h%20%3D%20%20%5Cfrac%7B1568%7D%7B40%7D%20%20%3D%20%20%5Cfrac%7B196%7D%7B5%7D%20%20%5C%5C%20%20)
We have the final answer as
<h3>39.2 m</h3>
Hope this helps you
You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be
<span>1 km = 0.6214 miles </span>
<span>1000 m = 1 km </span>
<span>60 seconds = 1 minute </span>
<span>60 minutes = 1 hour. </span>
<span>2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?</span>
Answer
given,
I is the loudness of sound
I = 10 Log₁₀ r
r is relative intensity
at when relative intensity is 10⁶
I = 60 dB
how much louder when 100 people would be talking together
I = 10 Log₁₀ r
I = 10 Log₁₀ (10⁶ x 100)
I = 10 Log₁₀ (10⁸)
I = 80 dB
hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.