The coefficient of friction between the soap and the floor is 0.081
If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.
We know that frictional force F = μN where μ = coefficient of friction between soap and floor.
So, μ = F/N
Since F = 40 N and N = W = 493 N,
μ = F/N
μ = 40 N/493 N
μ = 0.081
So, the coefficient of friction between the soap and the floor is 0.081
Learn more about coefficient of friction here:
brainly.com/question/13923375
Answer:2800000j
Explanation:
For us to know the kinetic energy of the vehicle,
Where m is the mass
And v is the velocity
Then, K.E=1/2mv^2
While, K.E=1/2×3500×40^2
Therefore, our answer will now be
K.E=2800000j
Answer:
ok confusion but we could figure it out right
Explanation:
<h3>dhdjhdndnd but its fine how was your day tho </h3>
Answer:
RE of Hydrogen = 6.47 x RE of Krypton
Explanation:
Actually the correct formula for comparing rate of effusion (RE) of two gases is:
RE of Gas A
------------------- = √ ( Molar mass of B / Molar mass of A)
RE of Gas B
You can designate which of the two gases you have (hydrogen and krypton) will be your gas A and gas B. So for this particular problem, let us make hydrogen as gas A and Krypton as gas B. So the equation becomes:
RE of Hydrogen
------------------------- = √ (Molar mass of Krypton / Molar mass of Hydrogen)
RE of Krypton
Get the molar masses of Hydrogen and Krypton in the periodi table:
RE of Hydrogen
------------------------- = √ (83.798 g/mol / 2 g/mol)
RE of Krypton
RE of Hydrogen
------------------------- = 6.47 ====> this can also be written as:
RE of Krypton
RE of Hydrogen = 6.47 x RE of Krypton
It means that the rate of effusion of Hydrogen gas will be 6.47 faster than the rate of effusion of Krypton gas. With the type of question you have, it doesn't matter which gases goes on your numerator and denominator. What's important is that you show the rate of effusion of a gas with respect to the other. But if that's concerns you the most, then take the gas which was stated first as your gas A and the latter as your gas B unless the problem tells you which one will be on top and which is in the bottom.
