Consider that I'm an italian student so I may unproperly use some terms.
Being Ug (Potential energy on earth) = m·g·h
Where "m" in this case is the mass of the ball, "g" the average value of gravitation constant on Earth's surface and "h" the height reached by the ball.
The solution to the first question would be : Ug = mgh = 40[kg] · 10 [m/s²] · 5[m] = 2000 [Joule]
In order to find the value of its K (kinetic energy) and without considering air resistance, you can use the concept of mechanical conservation of energy that confirms : U(start) + K(start) = U(final) + K(final).
So since you want to find its K(final) you can isolate it in the law and it gives you:
K(final) = U(start) + K(start) - U(final)
Consider that in halfway fall its Ug(final) value is mgh = 40[kg] · 10[m/s²] · 2.5[m] = 1000[Joule]
Now you just need to change these elements into their values(consider that K(start) is 0:
K(final) = 2000[Joule] + 0[Joule] - 1000[Joule] = 1000[J]
Hope I explained it well.
Answer:
Explanation:
Attach is A and B
c) No, because 2.6 is outside the range of the data values for the explanatory variable.
d) H0: β1 =0 and Ha: β1 > 0
Answer:
specifically 30º west of the north
Explanation:
This exercise can be solved using the composition of vectors, let's call the speed of the river that goes east (v₁ = 8 mi / h) to the speed of the boat (v₂ = 16 mi / h) and the desired speed from south to north (v₃ =?). Look at the attachment to be clear about the vectors
Let's use the Pythagorean triangle
v₃² = v₁² + v₂²
Let's calculate v₂
v₂² = v₃² - v₁²
v₂ = √ (16² - 8²)
v₂ = 13.86 mi / h
To find the angle we use trigonometry
sin θ = v1 / v3
θ = sin⁻¹ (v1 / v3)
θ = sin⁻¹ (8/16)
θ = 30º
The boat must be pointed at 30º with respect to the south-north direction, specifically 30º west of the north