Answer:
kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Explanation:
Given the data in the question;
we know that;
Kinetic energy = 1/2.mv²
given that mass of the object is doubled; m1 = 2m
speed is halved; v1 = V/2
Now, New kinetic energy will be; 1/2.m1v1²
we substitute
Kinetic Energy = 1/2 × 2m × (v/2)²
Kinetic Energy = 1/2 × 2m × (v²/4)
Kinetic Energy = 1/2 × m × (v²/2)
Kinetic Energy = 1/2 [ 1/2mv² ]
Kinetic Energy = 1/2 [ KE ]
Therefore; kinetic energy will change by a factor of 1/2
Option C) 1/2 is the correct answer
Answer:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
A chart that organizes all elements by their atomic number.
Answer:
The distance is 
Explanation:
From the question we are told that
The distance from the conversation is 
The intensity of the sound at your position is 
The intensity at the sound at the new position is 
Generally the intensity in decibel is is mathematically represented as
![\beta = 10dB log_{10}[\frac{d}{d_o} ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%2010dB%20log_%7B10%7D%5B%5Cfrac%7Bd%7D%7Bd_o%7D%20%5D)
The intensity is also mathematically represented as
So
![\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]](https://tex.z-dn.net/?f=%5Cbeta%20%20%3D%20%2010dB%20%2A%20%20log_%7B10%7D%5B%5Cfrac%7BP%7D%7BA%2A%20d_o%7D%20%5D)
=> ![\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cbeta%7D%7B10%7D%20%20%3D%20%20log_%7B10%7D%20%5B%5Cfrac%7BP%7D%7BA%20%28l_o%29%7D%20%5D)
From the logarithm definition
=> 
=> ![P = A (d_o ) [10^{\frac{\beta }{ 10} } ]](https://tex.z-dn.net/?f=P%20%3D%20%20A%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta%20%7D%7B%2010%7D%20%7D%20%5D)
Here P is the power of the sound wave
and A is the cross-sectional area of the sound wave which is generally in spherical form
Now the power of the sound wave at the first position is mathematically represented as
![P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]](https://tex.z-dn.net/?f=P_1%20%3D%20%20A_1%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_1%20%7D%7B%2010%7D%20%7D%20%5D)
Now the power of the sound wave at the second position is mathematically represented as
![P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20A_2%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_2%20%7D%7B%2010%7D%20%7D%20%5D)
Generally power of the wave is constant at both positions so
![A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]](https://tex.z-dn.net/?f=A_1%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_1%20%7D%7B%2010%7D%20%7D%20%5D%20%20%3D%20A_2%20%28d_o%20%29%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_2%20%7D%7B%2010%7D%20%7D%20%5D)
![4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]](https://tex.z-dn.net/?f=4%20%5Cpi%20r_1%20%5E2%20%20%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_1%20%7D%7B%2010%7D%20%7D%20%5D%20%20%3D%204%20%5Cpi%20r_2%20%5E2%20%20%20%5B10%5E%7B%5Cfrac%7B%5Cbeta_2%20%7D%7B%2010%7D%20%7D%20%5D)
![r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}](https://tex.z-dn.net/?f=r_2%20%3D%20%20%5Csqrt%7Br_1%20%5E2%20%5B%5Cfrac%7B10%5E%7B%5Cfrac%7B%5Cbeta_1%7D%7B10%7D%20%7D%7D%7B%2010%5E%7B%5Cfrac%7B%5Cbeta_2%7D%7B10%7D%20%7D%7D%20%5D%7D)
substituting value
![r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}](https://tex.z-dn.net/?f=r_2%20%3D%20%20%20%5Csqrt%7B%2024%5E2%20%5B%5Cfrac%7B10%5E%7B%5Cfrac%7B%2040%7D%7B10%7D%20%7D%7D%7B10%5E%7B%5Cfrac%7B80%7D%7B10%7D%20%7D%7D%20%5D%7D)
Answer:
Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
Energy can also be defined as the ability to do work.
