Answer:
because both petrol and diesel are oil
Explanation:
oil floats on water that's why if we will try to extinguish with water so the fire will float on <u>water</u>
<u>hope</u><u> </u><u>u</u><u> </u><u>like</u><u> </u><u>my</u><u> </u><u>ans</u><u>wer</u><u> </u>
<u>pl</u><u>ease</u><u> </u><u>mar</u><u>k</u><u> </u><u>methe</u><u> </u><u>brainest</u>
Answer:
Q=185.84C
Explanation:
We have to take into account the integral
![Q=\int \rho dV](https://tex.z-dn.net/?f=Q%3D%5Cint%20%5Crho%20dV)
In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4
![Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx](https://tex.z-dn.net/?f=Q%3D%5Cint_%7B-2%7D%5E2%5Cint_%7B-%5Csqrt%7B4-x%5E2%7D%7D%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx)
but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!
When more than one force acts upon an object, the vector sum of these forces is the resultant force. When the resultant force on an object is zero, it will remain at rest if it is at rest, or continue to move in a straight line at a constant velocity if it is in motion. ... That is, there is zero acceleration.
Answer:
normal force = 10 N
Explanation:
Given data
frictional force = 0.400 N
coefficient of kinetic friction = 0.04
Solution
we get here normal force that is express as
normal force =
............1
put here value and we will get value
normal force =
solve it we get
normal force = 10 N