Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The coefficient of kinetic friction between the sled and the snow is 0.0134
Explanation:
Given that:
M = mass of person = 52 kg
m = mass of sled = 15.2 kg
U = initial velocity of person = 3.63 m/s
u = initial velocity of sled = 0 m/s
After collision, the person and the sled would move with the same velocity V.
a) According to law of momentum conservation:
Total momentum before collision = Total momentum after collision
MU + mu = (M + m)V

Substituting values:

The velocity of the sled and person as they move away is 2.81 m/s
b) acceleration due to gravity (g) = 9.8 m/s²
d = 30 m
Using the formula:

The coefficient of kinetic friction between the sled and the snow is 0.0134
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,

. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:

Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:

Now we can use the following relationship to find the distance covered by the skier before stopping, S:

where

is the final speed of the skier and

is the initial speed. Substituting numbers, we find:
The velocity is 14 m/s
The parameters given on the question are
mass= 0.060 kg
kinetic energy= 5.9 joules
K.E= 1/2mv²
5.9= 1/2 × 0.060 × v²
5.9= 0.5 × 0.060v²
5.9= 003v²
v²= 5.9/0.03
v²= 196.66
v= √196.66
v= 14 m/s
Hence the velocity of the egg before it strikes the ground is 14 m/s
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Answer:
Answer = 480000 joule.
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