P=4800kgm/s
As
p=mΔv
where p is momentum, m is mass and v is velocity
Given values is
m =1200kg
Δv= 17m/s-13m/s=4m/s
Now
p=mΔv
p=(1200kg)*(4m/s)
p=4800kgm/s
Average speed is the ratio of total distance moved by Chi in total time interval
So here we will have
Total distance = 100 m + 400 m
![d = 500 m = 0.5 km](https://tex.z-dn.net/?f=d%20%3D%20500%20m%20%3D%200.5%20km)
Total time taken = 5 min + 15 min = 20 min
![T = \frac{20}{60} = \frac{1}{3} hour](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B20%7D%7B60%7D%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20hour)
now by the formula of average speed we know that
![v_{avg} = \frac{d}{T}](https://tex.z-dn.net/?f=v_%7Bavg%7D%20%3D%20%5Cfrac%7Bd%7D%7BT%7D)
![v_{avg} = \frac{0.5 km}{1/3 h}](https://tex.z-dn.net/?f=v_%7Bavg%7D%20%3D%20%5Cfrac%7B0.5%20km%7D%7B1%2F3%20h%7D)
![v_{avg} = 1.5 km/h](https://tex.z-dn.net/?f=v_%7Bavg%7D%20%3D%201.5%20km%2Fh)
so average speed will be 1.5 km/h
Answer:
a
![Q = 5.34 *10^{19} \ J](https://tex.z-dn.net/?f=Q%20%20%3D%20%205.34%20%2A10%5E%7B19%7D%20%5C%20%20J)
b
![T = 0.445 * 365 = 162. 413 \ days](https://tex.z-dn.net/?f=T%20%20%3D%200.445%20%2A%20365%20%3D%20%20162.%20413%20%20%5C%20days)
Explanation:
From the question we are told that
The area of Manhattan is ![a_k = 87.46 *10^{6} \ m^2](https://tex.z-dn.net/?f=a_k%20%20%3D%20%2087.46%20%2A10%5E%7B6%7D%20%5C%20%20m%5E2)
The area of the ice is ![a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2](https://tex.z-dn.net/?f=a_i%20%20%3D%20%204%2A%20%2087.46%20%2A10%5E%7B6%20%7D%20%3D%203.498%20%2A10%5E%7B8%7D%5C%20m%5E2)
The thickness is ![t = 500 \ m \\](https://tex.z-dn.net/?f=t%20%20%3D%20%20500%20%5C%20m%20%20%5C%5C)
Generally the volume of the ice is mathematically represented is
![V = a_i * t](https://tex.z-dn.net/?f=V%20%20%3D%20%20a_i%20%2A%20%20t)
substituting value
![V = 500 * 3.498*10^{8}](https://tex.z-dn.net/?f=V%20%20%3D%20%20500%20%2A%203.498%2A10%5E%7B8%7D)
![V = 1.75 *10^{11}\ m^3](https://tex.z-dn.net/?f=V%20%20%3D%20%201.75%20%2A10%5E%7B11%7D%5C%20m%5E3)
Generally the mass of the ice is
![m_i = \rho_i * V](https://tex.z-dn.net/?f=m_i%20%20%3D%20%20%5Crho_i%20%20%2A%20%20V)
Here
is the density of ice the value is ![\rho _i = 916.7 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho%20_i%20%20%3D%20%20916.7%20%5C%20kg%2Fm%5E3)
=> ![m_i = 916.7 * 1.75*10^{11}](https://tex.z-dn.net/?f=m_i%20%20%3D%20%20916.7%20%20%20%2A%20%20%201.75%2A10%5E%7B11%7D)
=> ![m_i = 1.60 *10^{14} \ kg](https://tex.z-dn.net/?f=m_i%20%20%3D%20%201.60%20%2A10%5E%7B14%7D%20%5C%20%20kg)
Generally the energy needed for the ice to melt is mathematically represented as
![Q = m _i * H_f](https://tex.z-dn.net/?f=Q%20%20%3D%20%20m%20_i%20%20%2A%20H_f)
Where
is the latent heat of fusion of ice and the value is ![H_f = 3.33*10^{5} \ J/kg](https://tex.z-dn.net/?f=H_f%20%20%3D%20%203.33%2A10%5E%7B5%7D%20%5C%20%20J%2Fkg)
=> ![Q = 1.60 *10^{14} * 3.33*10^{5}](https://tex.z-dn.net/?f=Q%20%20%3D%20%201.60%20%2A10%5E%7B14%7D%20%2A%20%203.33%2A10%5E%7B5%7D)
=> ![Q = 5.34 *10^{19} \ J](https://tex.z-dn.net/?f=Q%20%20%3D%20%205.34%20%2A10%5E%7B19%7D%20%5C%20%20J)
Considering part b
We are told that the annual energy consumption is ![G = 1.2*10^{20 } \ J / year](https://tex.z-dn.net/?f=G%20%20%3D%20%201.2%2A10%5E%7B20%20%7D%20%5C%20J%20%20%2F%20year)
So the time taken to melt the ice is
![T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}](https://tex.z-dn.net/?f=T%20%20%3D%20%20%5Cfrac%7B%205.34%20%2A10%5E%7B19%7D%7D%7B%201.2%20%2A10%5E%7B20%7D%7D)
![T = 0.445 \ years](https://tex.z-dn.net/?f=T%20%20%3D%200.445%20%5C%20years)
converting to days
![T = 0.445 * 365 = 162. 413 \ days](https://tex.z-dn.net/?f=T%20%20%3D%200.445%20%2A%20365%20%3D%20%20162.%20413%20%20%5C%20days)
That's false. Displacement would be (r2 - r1) .
Answer:
7.3km/hr
Explanation:
v=d/t=4.5km/0.62hrs=7.3km/hr