Max preassure = force / min area
= 3N / 0.1 x 0.05
= 600N/m(squared)
Copy off of the picture below itll help better, its what someone sent me when i asked this question
<u>Answer:</u>
2.39 kg
<u>Explanation:</u>
There is conservation of momentum here in this problem so we will use the following problem:
where the mass of the student 
is 48.5 kg,
the mass of the skateboard 
is kg,
the initial speed of the student 
is 4.25 m/s; and
the speed of the student and skateboard 
is 4.05 m/s.
So substituting the given values in the above formula to get:
Therefore, the mass of the skateboard is 2.39 kg.
Answer:
1. They both uses same energy
2. The 6 kg ball requires more power than 3kg ball
Explanation:
Sample 1
m = 3kg
g= 10m/s^2
h = 2m
t = 2secs
W = mgh = 3 x 10 x 2 = 60J
P= w/t = 60/2 = 30watts
Sample 2
m = 6kg
g= 10m/s^2
h = 1m
t = 1sec
W = mgh = 6 x 10 x 1 = 60J
P= w/t = 60/1 = 60watts
They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball
Answer;
Q = 359.2-J
Explanation;
Given that;
Constants for mercury at 1 atm
Heat capacity of Hg(l) is 28.0 J/(mol*K)
melting point is 234.32 K
Enthalpy of fusion is 2.29 kJ/mol
17.7-g Hg / 200.6g/mol = 0.0882 mol Hg;
°C + 273 = 298 K;
2.29-kJ/mol = 2290-J/mol
Q = (m x ΔT x Cp) + (m x Hf)
Q = 0.0882-mol x (298 - 234.32) x 28.0-J/mol*K) + (0.0882-mol x 2290-J/mol)
Q = 157.26-J + 201.978-J
Q = 359.2-J
Q=359-J (3 sig fig allowed due to 17.7-g given in problem)
Answer:
L₁ = W×L / w
Explanation:
The scenario is shown in the image below.
<u>At the pivot point, the torque acting on this point must be zero so that there will be easy back and forth without the heavier child.</u>
<u>Torque created by lighter child + Torque created by the heavier child = 0</u>
Thus,
According to the axis system, the heavier child is left to the pivot (origin), so,
W×(- L ) + w× L₁ = 0
So,
<u>L₁ = W×L / w</u>
