Answer:
A plant
Explanation:
because animals don't have cell walls, and fungus and bacteria dont have chloroplasts
A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.
<u>Explanation:</u>
The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.
It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.
If the resistance of the Air is ignored, we can use the theory given by Galileo in which he warned that the thermal velocity of a body in free fall was given by

Where
g = Gravitational acceleration
t = time
As we can see the speed of objects in free fall is indifferent to the position that is launched (as long as the resistance of the air is ignored) or its mass.
Both bodies will end with the same thermal speed.
Ruff's image is 50m behind the mirror surface and the image is also 3m tall.
This is because it is a plane mirror.
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two.
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂)
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) )
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂
But I suppose we ought to kick that idea around a bit.
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D.
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁²
Differentiate with respect to d₁
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero.
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that
d₁ = d₂ = ½D so
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)