Answer:
<em>Correct choice: b 4H</em>
Explanation:
<u>Conservation of the mechanical energy</u>
The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):
E = U + K
The GPE is calculated as:
U = mgh
And the kinetic energy is:
Where:
m = mass of the object
g = gravitational acceleration
h = height of the object
v = speed at which the object moves
When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:
When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:
Since the energy is conserved, U1=U2
![\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%20%20%20%5Cqquad%5Cqquad%20%5B1%5D)
For the speed to be double, we need to drop the snowball from a height H', and:
Operating:
![\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%27%3D4%5Cfrac%7B1%7D%7B2%7Dm%28v%29%5E2%20%5Cqquad%5Cqquad%20%5B2%5D)
Dividing [2] by [1]
Simplifying:
Thus:
H' = 4H
Correct choice: b 4H
Answer:
b
c
e
h
Explanation:
Note that the swing direction was not giving in the question and direction could be sideways (in a turn) or in a track or both
The question show something in common ...acceleration
So let's look at the statements and pick the correct ones
a is false while b is correct as the train is accelerating
c is correct. The train is accelerating even thou the speed could not be ascertained
d is false and not feasible as the train is accelerating
e is true as the train maybe moving at a constant speed in a circle
f is false. This could be constant velocity in a circle. Same as g (false)
h is true. It's accelerating
Answer:
The light used has a wavelenght of 4.51×10^-7 m.
Explanation:
let:
n be the order fringe
Ф be the angle that the light makes
d is the slit spacing of the grating
λ be the wavelength of the light
then, by Bragg's law:
n×λ = d×sin(Ф)
λ = d×sin(Ф)/n
λ = (3.2×10^-4 cm)×sin(25.0°)/3
= 4.51×10^-5 cm
≈ 4.51×10^-7 m
Therefore, the light used has a wavelenght of 4.51×10^-7 m.
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
