Answer:
31.25 m
25m/sec
Explanation:
Given :-
Time = 5sec
V = 0 (in going up)
U = 0 (in comming down)
Find :-
H and U by which it is thrown up
Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .
We know that ,
V = U + gt
0 = U - 10*2.5
U = 25 m/sec
Also,
V² = U² +2gs
0 = 625 - 20s
s = 625/20 = 31.25 m
Answer:
609547.12 Pa ≈ 6.10×10^5 Pa
Explanation:
Step 1:
Data obtained from the question. This include the following:
Force (F) = 49.8 N
Radius (r) = 0.00510 m
Pressure (P) =..?
Step 2:
Determination of the area of the head of the nail.
The head of a nail is circular in nature. Therefore, the area is given by:
Area (A) = πr²
With the above formula we can obtain the area as follow:
Radius (r) = 0.00510 m
Area (A) =?
A = πr²
A = π x (0.00510)²
A = 8.17×10^-5 m²
Therefore the area of the head of the nail is 8.17×10^-5 m²
Step 3:
Determination of the pressure exerted by the hammer.
This is illustrated below:
Force (F) = 49.8 N
Area (A) = 8.17×10^-5 m²
Pressure (P) =..?
Pressure (P) = Force (F) /Area (A)
P = F/A
P = 49.8/8.17×10^-5
P = 609547.12 N/m²
Now, we shall convert 609547.12 N/m² to Pa.
1 N/m² = 1 Pa
Therefore, 609547.12 N/m² = 609547.12 Pa.
Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa
Answer:
Explanation:
As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have
Work done by friction force = Loss in mechanical energy
so we will have
here we know that
Initial compression in the spring is given as
now from above equation
Steel is more dense because it’s heavy while water is light
I think it 32, but i’m not sure
