To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.
By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore
Here,
m = mass
g =Gravitational energy
The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore
Remember the expression for which you can determine the relationship between mass, volume and density, in which
In this case the density would be that of the object, replacing
Since the displaced volume of water is 0.429 we will have to
The density of water under normal conditions is 
, so
The density of the object is 
Answer:
23. 4375 m
Explanation:
There are two parts of the rocket's motion
1 ) accelerating (assume it goes upto h1 height )
using motion equations upwards
Lets find the velocity after 2.5 seconds (V1)
V = U +at
V1 = 0 +5*2.5 = 12.5 m/s
2) motion under gravity (assume it goes upto h2 height )
now there no acceleration from the rocket. it is now subjected to the gravity
using motion equations upwards (assuming g= 10m/s² downwards)
V²= U² +2as
0 = 12.5²+2*(-10)*h2
h2 = 7.8125 m
maximum height = h1 + h2
= 15.625 + 7.8125
= 23. 4375 m
Here stress is parallel to the surface of the body. So it's a Shear stress.
Answer:
20 meters.
Explanation:
In the graph, the x-axis (the horizontal axis) represents the time, while the y-axis (the vertical axis) represents the distance.
If we want to find the distance covered in the first T seconds, you need to find the value T in the horizontal axis.
Once you find it, we draw a vertical line, in the point where this vertical line touches the graph, we now draw a horizontal line. This horizontal line will intersect the y-axis in a given value. That value is the total distance travelled by the time T.
In this case, we want to find the total distance that David ran in the first 4 seconds.
Then we need to find the value 4 seconds in the horizontal axis. Now we perform the above steps, and we will find that the correspondent y-value is 20.
This means that in the first 4 seconds, David ran a distance of 20 meters.
Go and click to the invitation bar and you can find an option written as " search friends " . Then it's easy to find that unknown user if you're pretty fond with his/her username and DP ( display picture ).
