Why are many mitochondria needed in cells that move, like muscle cells?
1 answer:
You might be interested in
Answer:
The weight of the object X is approximately 3.262 N (Acting downwards)
The weight of the object Y is approximately 8.733 N (Acting downwards)
Explanation:
The question can be answered based on the principle of equilibrium of forces
The given parameters are;
The weight of Z = 12 N (Acting downwards)
The weight of the pulleys = Negligible
From the diagram;
The tension in the in the string attached to object Z = The weight of object Z = 12 N
The tension in the in the string attached to object X = The weight of the object X
The tension in the in the string attached to object Y = The weight of the object Y
Given that the forces are in equilibrium, we have;
The sum of vertical forces acting at a point, = 0
Therefore;
Where;
= The weight of object Z = 12 N
= 12 N
= The vertical component of tension, T₂ = T₂ × sin(24°)
∴ = T₂ × sin(156°)
Similarly;
= T₃ × sin(50°)
From , and
= 12 N, we have;
12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)
Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0
Therefore at point B, we have;
T₁ₓ + T₂ₓ + T₃ₓ = 0
The tension force, T₁, only has a vertical component, therefore;
∴ T₁ₓ = 0
∴ T₂ₓ + T₃ₓ = 0
T₂ₓ = -T₃ₓ
T₂ₓ = T₂ × cos(156°)
T₃ₓ = T₃ × cos(50°)
From T₂ₓ = -T₃ₓ, we have;
T₂ × cos(156°) = - T₃ × cos(50°)...(2)
Making T₃ the subject of equation (1) and (2) gives;
Making T₃ the subject of equation in equation (1), we get;
12 = -(T₂ × sin(156°) + T₃ × sin(50°))
∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))
Making T₃ the subject of equation in equation (2), we get;
T₂ × cos(156°) = - T₃ × cos(50°)
∴ T₃ = T₂ × cos(156°)/(-cos(50°))
Equating both values of T₃ gives;
(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))
-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))
∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283
∴ T₂ ≈ -8.02 N
From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;
T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717
∴ T₃ ≈ -11.4 N
The weight of the object X = -T₂ × sin(156°)
∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N
The weight of the object X ≈ 3.262 N (Acting downwards)
The weight of the object Y = -(T₃ × sin(50°))
∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N
The weight of the object Y ≈ 8.733 N (Acting downwards)
Answer:
104.6 m
Explanation:
The motion of the car is a projectile motion, consisting of:
- A horizontal motion at constant speed
- A vertical motion at constant acceleration (acceleration of gravity)
We know that:
The initial horizontal velocity of the car is , and this is constant during the entire motion, as there are no forces in the horizontal direction
is the horizontal distance travelled by the car
Therefore, we can find the total time of flight of the car:
Now we analyze the vertical motion, which is a free fall motion, so we can use the suvat equation:
where
s is the vertical displacement (= the height of the cliff)
u = 0 is the initial vertical velocity
t = 4.62 s is the time of flight
is the acceleration of gravity
Solving for s, we find the height of the cliff: