2.5 cm 1 meter 1 Year
_______ x ________ x ________ = 0.002 m/month
Year 100 cm 12 mo.s
Answer:
825 kgm⁻³
Explanation:
ρ = density of wood = ?
ρ' = density of water = 1000 kgm⁻³
V = volume of wood = 10 x 4 x 2 = 80 cm³ = 80 x 10⁻⁶ m³
V' = Volume of water displaced = 10 x 4 x 1.65 = 66 cm³ = 66 x 10⁻⁶ m³
Using equilibrium of force in vertical direction
Force of buoyancy = Weight of the wood
ρ' V' g = ρ V g
ρ' V' = ρ V
(1000) (66 x 10⁻⁶) = ρ (80 x 10⁻⁶)
ρ = 825 kgm⁻³
First draw both axes then draw a vector of (theoretical) length 3.5 along the negative x axes.
Then draw a "8.2" vector from that endpoint all the way up, 60 degrees from that point (90-30) This is because you make a 30 degree angle with the current vertical and then you will see that the angle in the triangle we are making is 90 degrees minus that 30 so we come up ith 60 degrees.
You should make the 8.20 vector higher that the orginal 3.5 m vector starting point.
Next we will finally draw a 15 m vector from the endpoint of the 8.2 m vector that goes above the original one and to the left. This one should be drawn really far.
We will solve for the sides. The first triangle we make is 30-60-90, so we must solve fot the vertical side which is just:
8.2/2 = 4.1 m
Now solve for the opposite side using 4.1(sqrt(3)) = 7.1 m. So now we have enough to solve for the other triangle that the 15m line makes.
the top side is 15 - 7.1 = 7.9 m
while the right side is 4.1 - 3.5 = 0.6 m
The resultant displacement is the last side:
√(7.9^2) + (0.6^2) = 7.9 m (ANSWER)
To find the direction simply use:
arcTan(y/x) = arcTan(0.6/7.9) = 4.3 degrees north of west (ANSWER 2)
Explanation:
The observation from the gold foil experiment by Rutherford was that, while most of the alpha particles went through the foil almost unaffected, a small fraction of the particles were deflected in directions with large angles away from the original path (including bouncing straight back). This was unexpected and led to a complete revision of the model (as relates to your part 1).
The nuclear model proposes a spacially small but massively charged nucleus. Due to its small size, most alpha particles will pass through the atoms of the gold foil unaffected (they "miss" the small nucleus). But some of them will come very close to the nucleus and those will be deflected strongly by its charge. This is one of the main arguments explaining the observation and speaking for the validity of the nuclear model.
Answer:
v = 60 m/s
Explanation:
It is given that,
A wave is represented by the equation :
We need to find the velocity of the wave
The general equation of a wave is given by :
....(1)
Equation (1) can be written as :
...(2)
If we compare equation (1) and (2) we get :
The velocity of a wave is given by :
So, the velocity of the wave is 60 m/s.