Answer:
F = -307.4 N
Explanation:
It is given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 60 m/s
Final speed of the baseball, 
Time of contact, 
(a) It is assumed to find the horizontal component of average force. It is given by :
F = -307.4 N
So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.
Answer:
a)V= 0.0827 m³
b)P=181.11 x 10² N/m²
Explanation:
Given that
m = 81.5 kg
Density ,ρ = 985 kg/m³
As we know that
Mass = Volume x Density
81.5 = V x 985
V= 0.0827 m³
The force exerted by weight = m g
F= m g= 81.5 x 10 = 815 N ( Take ,g= 10 m/s²)
Area ,A= 4.5 x 10⁻² m²
The Pressure P


P=181.11 x 10² N/m²
Answer:
Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here
Explanation:
As we know that zinc reacts with copper sulfate
so the reaction is given as

so here we have




Now total mass of reactant is given as

Mass of the product is given as

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here
The S.I. unit for the measure of the pressure is the Pascal (Pa). 1 Pascal corresponds to

We can convert the number given by the problem into Pascal:

And since

, we have