Answer:
The mass of the mud is 3040000 kg.
Explanation:
Given that,
length = 2.5 km
Width = 0.80 km
Height = 2.0 m
Length of valley = 0.40 km
Width of valley = 0.40 km
Density = 1900 Kg/m³
Area = 4.0 m²
We need to calculate the mass of the mud
Using formula of density
![\rho=\dfrac{m}{V}](https://tex.z-dn.net/?f=%5Crho%3D%5Cdfrac%7Bm%7D%7BV%7D)
![m=\rho\times V](https://tex.z-dn.net/?f=m%3D%5Crho%5Ctimes%20V)
Where, V = volume of mud
= density of mud
Put the value into the formula
![m=1900\times4.0\times0.40\times10^{3}](https://tex.z-dn.net/?f=m%3D1900%5Ctimes4.0%5Ctimes0.40%5Ctimes10%5E%7B3%7D)
![m =3040000\ kg](https://tex.z-dn.net/?f=m%20%3D3040000%5C%20kg)
Hence, The mass of the mud is 3040000 kg.
1
2 4. 5 4 5
+3 0 7. 3 0 0
——————
3 3 1 8 4 5
line up the decimal points and add.
hope this helps!
Answer:
here
Explanation:
There are two forces acting upon the skydiver - gravity (down) and air resistance (up). The force of gravity has a magnitude of m•g = (72 kg) •(9.8 m/s/s) = 706 N. ... a 3.25-kg object rightward with a constant acceleration of 1.20 m/s/s if the force of ... of 33.8 kg, how far (in meters) will it move in 1.31 seconds, starting from rest?
Answer:
Force, ![F=2.27\times 10^4\ N](https://tex.z-dn.net/?f=F%3D2.27%5Ctimes%2010%5E4%5C%20N)
Explanation:
Given that,
Mass of the bullet, m = 4.79 g = 0.00479 kg
Initial speed of the bullet, u = 642.3 m/s
Distance, d = 4.35 cm = 0.0435 m
To find,
The magnitude of force required to stop the bullet.
Solution,
The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :
![F.d=\dfrac{1}{2}m(v^2-u^2)](https://tex.z-dn.net/?f=F.d%3D%5Cdfrac%7B1%7D%7B2%7Dm%28v%5E2-u%5E2%29)
Finally, it stops, v = 0
![F.d=-\dfrac{1}{2}m(u^2)](https://tex.z-dn.net/?f=F.d%3D-%5Cdfrac%7B1%7D%7B2%7Dm%28u%5E2%29)
![F=\dfrac{-mu^2}{2d}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B-mu%5E2%7D%7B2d%7D)
![F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B-0.00479%5Ctimes%20%28642.3%29%5E2%7D%7B2%5Ctimes%200.0435%7D)
F = -22713.92 N
![F=2.27\times 10^4\ N](https://tex.z-dn.net/?f=F%3D2.27%5Ctimes%2010%5E4%5C%20N)
So, the magnitude of the force that stops the bullet is ![2.27\times 10^4\ N](https://tex.z-dn.net/?f=2.27%5Ctimes%2010%5E4%5C%20N)