Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
The reason you only see fossils in sedimentary rock is that these set of rocks are formed in much lower pressure and temperature, compared to the other types of rocks like igneous rocks.
ans will be 1500006.15
= 1.5*10^6
we move the decimal point to the left six digits
Answer:
The heat flux between the surface of the pond and the surrounding air is<em> 60 W/</em>
<em> </em>
Explanation:
Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;
q = Q/A ...............................1
since we are working with the convectional heat transfer coefficient equation 1 become;
q = h (
) ........................2
where q is the heat flux;
Q is the heat energy that will be transferred;
h is the convectional heat coefficient = 20 W/.K;
is the surface temperature = 
C 23°C + 273.15 = 296.15 K;
is the surrounding temperature = 
C = 20°C + 273.15 = 293.15 K;
The values are substituted into equation 2;
q = 20 W/.K ( 296.15 K - 293.15 K)
q = 20 W/.K ( 3 K)
q = 60 W/
Therefore the heat flux between the surface of the pond and the surrounding air is 60 W/
