Answer:
a) h'= 5/7 h
, b) h ’= h
Explanation:
Let's use energy conservation for this exercise
Starting point. Upper left side
Em₀ = mg h
Final point. Lower left side
Emf = K = ½ m v² + ½ I w²
Em₀ = emf
mgh = ½ m v² + ½ I w²
Angular and linear velocity are related
v = r w
w = v / r
The moment of inertia of the marble that we take as a solid sphere is
I = 2/5 m r²
We substitute
m g h = ½ m v² + ½ 2/5 m r² (v / r)²
g h = ½ v2 (1 + 2/5)
v = √(g h 10/7)
This is the speed at the bottom of the bowl
Now let's apply energy conservation to the right side
a) right side if rubbing
Em₀ = K
Emf = U = mg h’
½ m v² = mg h’
h’= ½ (g h 10/7) / g
h'= 5/7 h
b) right side with rubbing
Em₀ = K
Emf = K + U = -½ I w² + m g h
Emf = -½ 2/5 m r² v² / r² + m gh
Em₀ = emf
½ v² = -1/5 v² + g h’
h’= (1/2 +1/5) (gh 10/7) / g
h ’= h
c) When there is friction, an energy of rotation is accumulated that must be dissipated, by local it goes higher
A compression is where the particles in the air are closer together, so where those black lines are closer together.
A rarefaction is the opposite, where they're spread out.
The wavelength is the distance between two compressions or rarefactions (i.e. two peaks or troughs on a graph), therefore thats the horizontal arrows.
The amplitude is the distance from the centre of the wave to the peak or trough, so that is the vertical distance on the diagram.
Answer:
40·919 m
Explanation:
Initial velocity of the arrow = 46 m/s
Angle at which it is thrown from horizontal = 38°
<h3>At the maximum height, the vertical component of velocity will be 0</h3>
Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s
From the formula
<h3>v² - u² = 2 × a × s</h3>
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Considering the formula in vertical direction and taking upward direction as positive
v = 0
u = 28·32 m/s
a = - g = - 9·8 m/s²
Let s be the maximum height
- 28·32² = - 2 × 9.8 × s
⇒ s = 40·919 m
∴ The arrow will go 40·919 m high
Answer:
Option C. The force between them would be 4 times larger than with the
initial masses.
Explanation:
To know which option is correct, we shall determine the force of attraction between the two masses when their masses are doubled. This can be obtained as follow:
From:
F = GMₐM₆/ r²
Keeping G/r² constant, we have
F₁ = MₐM₆
Let the initial mass of both objects to be m
F₁ = MₐM₆
F₁ = m × m
F₁ = m²
Next, let the masses of both objects doubles i.e 2m
F₂ = MₐM₆
F₂ = 2m × 2m
F₂ = 4m²
Compare the initial and final force
Initial force (F₁) = m²
Final (F₂) = 4m²
F₂ / F₁ = 4m² / m²
F₂ / F₁ = 4
F₂ = 4F₁ = 4m²
From the above illustrations, we can see that when the mass of both objects doubles, the force between them would be 4 times larger than with the
initial masses.
Thus, option C gives the correct answer to the question.
