Gamma Rays Hope This Helps
First let's find the time it takes for the first ball to land:
Acceleration is a=-g so vertical velocity is V=-gt + V1sin(30).
Position is thus
S=(-1/2)gt^2 +V1t sin(30).
Solving for t gives
t=2V1sin(30)/g
The second ball has the same position function except for the new velocity, which is given by
V2=2V1. Putting this in and solving for t2 gives
t2=4V1sin(30)/g.
It takes twice as long for the second ball to land on the ground.
The horizontal distance of ball 1 is S1 = V1t cos(30). Again we look at ball 2's distance by substituting V2=2V1 and get
S2 = 2V1t2 cos(30).
Note here I put in t2 since it will fly for that amount of time. But we already saw that
t2 = 2t1
So S2=4V1 cos(30)
That is the second ball goes 4 times further than the first one. This is because it is going twice as fast along both the horizontal and the vertical. It moves horizontally twice as fast for twice as long.
Thomson proposed the "plum pudding" model - randomly scattered positive charges and negative charges held in a sphere
Rutherford disagreed with Thomson with his gold foil experiment - proposed that at the centre of an atom was a dense, positive mass and around it were very small negative charges called electrons
Answer: B. irregular galaxy
Explanation: "An irregular galaxy is a galaxy that does not have a distinct regular shape, unlike a spiral or an elliptical galaxy."
<u>Given:</u>
Enthalpy of dissolution for NaOH = 44 kJ/mol
Mass of NaOH = 12.6 g
Mass of water = 250 g
Initial temperature of water T1 = 23 C
<u>To determine:</u>
The final temperature of water, T2
<u>Explanation:</u>
Molar mass of NaOH = 40 g/mol
# moles of NaOH = 12.6 g/ 40 g.mol-1 = 0.315 moles
Heat released during dissolution of 0.315 moles of NaOH=
= 44 * 0.315 = 13.86 kJ = 13860 J
Heat released by NaOH = heat absorbed by water
13860 = m*c*(T2-T1)
13860 = 250*4.18*(T2-23)
T2 = 36.26 C
Ans: The final temperature of water is 36.3 C