Answer:

Explanation:
From the exercise we know the final x distance, the angle which the projectile is being released and acceleration of gravity

From the equation of x-position we know that

Solving for 
(1)
Now, if we analyze the equation of y-position we got

At the end of the motion y=0

Knowing the equation for
in (1)


Solving for t
Now, we can solve (1)

Answer:
Let d be the density of fluid.
So , Initial reading of balance, F1 =30dg N
After the level reaches 50cm^3
Final reading of balance , F2 =50dg N
Given that difference between final and initial reading is 30g
i.e, F2 −F1
=30 g
⟹50dg−30dg=30g
⟹20dg=30g
⟹d=30g/20g
⟹d=1.5g/cm^3
So, density of fluid is 1.5g/cm^3
Answer:

Explanation:
The minimum total work is the work needed to counteract the work associated with the weight:




C does not represent a benefit, so that seems most logical.
A) they comprised.
B) at least ones listening
D) nothings wring with that
Answer:
All atoms of an element are identical. The atoms of different elements vary in size and mass. Compounds are produced through different whole-number combinations of atoms. A chemical reaction results in the rearrangement of atoms in the reactant and product compounds.
HOPE THIS HELPED!!!!!!!!!!!!!!!!!!! XDDDDDDDDDDD