Answer:
1.94 kg
Explanation:
we have given density of water = 1 gm/
the initial reading of the scale = 19 N
change in reading of scale =10 N
we have to find the mass of the block when the block is lowered in water
mass of the block when it is lowered in the water is given by
here g is the gravity of acceleration which value is given by 9.8
so the mass of the block =
Answer:
The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.
C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'
Explanation:
See attached picture.
Answer:
Pressure difference between Top and Bottom of the cylinder is given as
Explanation:
As we know that the force due to pressure is balanced by the weight of the cylinder
So we will have
so we have
so we have
so we have
Answer:
The shell will land 10.18m away from the buoy.
Explanation:
In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).
Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.
In order to do so we can use the following equation:
now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:
and solve for t:
now we can substitute the values:
t=3.19s
Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.
So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:
We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:
So we can substitute the given values:
Which yields:
so now we can use the A and B points to find by how far the shell missed the buoy:
Distance=329.18m-319m=10.18m
So the shell missed the buoy by 10.18m.
