Prokaryotes lack an organized nucleus and other membrane<span>-bound organelles. Prokaryotic DNA is found in a central part of the cell called the nucleoid. The cell wall of a prokaryote </span>acts<span> as an extra </span>layer<span> of protection helps maintain </span>cell shape<span>, and prevents dehydration.
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Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
The time taken for the two balls to hit each other is 8 s.
The given parameters:
- <em>Acceleration of the rocket, a = 2 m/s²</em>
- <em>Length of the chamber, s = 4 m</em>
- <em>Speed of the first ball, = V1 = 0.3 m/s</em>
- <em>Speed of the second ball, V2 = 0.2 m/s</em>
The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;

Thus, the time taken for the two balls to hit each other is 8 s.
Learn more about relative velocity here: brainly.com/question/17228388
it will move the object if you apply force, because of that it will stay the same with newton's law.
Answer:
a) m=20000Kg
b) v=0.214m/s
Explanation:
We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.
For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is,
, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as
, and the mass of the first and second coals as
and
respectively
We start with the transition between parts A and B, so we have:

Which means

And since we want the mass of the first coal thrown (
) we do:



Substituting values we obtain

For the transition between parts B and C, we can write:

Which means

Since we want the new final speed of the car (
) we do:

Substituting values we obtain
