A)<span>the production of bubbles by the plant when it is in darkness because in the dark the plant respires and when it does so it produces CO2 gas not O2</span>
Answer:
1300.80988 Nm
730.79206 kgm²
Explanation:
m = Mass of ladder = 23.8 kg
g = Acceleration due to gravity = 9.81 m/s²
L = Length of ladder = 9.08 m
F = Applied force = 260 N
= Angular acceleration = 1.78 rad/s²
I = Moment of inertia
Weight is given by
![W=mg\\\Rightarrow W=23.8\times 9.81\\\Rightarrow W=233.478\ N](https://tex.z-dn.net/?f=W%3Dmg%5C%5C%5CRightarrow%20W%3D23.8%5Ctimes%209.81%5C%5C%5CRightarrow%20W%3D233.478%5C%20N)
The center of gravity of the ladder lies at the center of the ladder
Torque will be
![\tau=-W\times \frac{L}{2}+F\times L\\\Rightarrow \tau=-233.478\times \frac{9.08}{2}+260\times 9.08\\\Rightarrow \tau=1300.80988\ Nm](https://tex.z-dn.net/?f=%5Ctau%3D-W%5Ctimes%20%5Cfrac%7BL%7D%7B2%7D%2BF%5Ctimes%20L%5C%5C%5CRightarrow%20%5Ctau%3D-233.478%5Ctimes%20%5Cfrac%7B9.08%7D%7B2%7D%2B260%5Ctimes%209.08%5C%5C%5CRightarrow%20%5Ctau%3D1300.80988%5C%20Nm)
The net torque acting on the ladder is 1300.80988 Nm
Torque is also given by
![\tau=I\alpha\\\Rightarrow I=\frac{\tau}{\alpha}\\\Rightarrow I=\frac{1300.80988}{1.78}\\\Rightarrow I=730.79206\ kgm^2](https://tex.z-dn.net/?f=%5Ctau%3DI%5Calpha%5C%5C%5CRightarrow%20I%3D%5Cfrac%7B%5Ctau%7D%7B%5Calpha%7D%5C%5C%5CRightarrow%20I%3D%5Cfrac%7B1300.80988%7D%7B1.78%7D%5C%5C%5CRightarrow%20I%3D730.79206%5C%20kgm%5E2)
The moment of inertia of the ladder is 730.79206 kgm²
0.32 km
How to reach the answer:
The formula below provides the tsunami's speed in response to the query.
s = 356√d
when the tsunami's speed is 200 km/h, the equation's "s" is changed to "200," and the equation of speed is then expressed as follows.
200 = 356√d
The equation can be changed to compute the depth of the water as,
√d = 200÷356
√d = 0.562
Now on squaring both sides:
(√d)² = (0.562)²
d = (0.562)² = 0.316 or 0.32
Therefore the approximate depth of water for a tsunami travelling at 200km/h is 0.32 km
Know more about speed here:
brainly.com/question/19930939
#SPJ4
An unstable isotope changes until it reaches a different element that is stable.
An unstable isotope changes until it reaches a different isotope of the same element that is stable.
i just did this on ed the other answer was wrong. these ones are right tho
Answer:
The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
Explanation:
Given;
moment of inertia of a skater with arms out,
= 3.1 kg.m²
moment of inertia of a skater with arms in,
= 0.9 kg.m²
inward angular speed,
= 4 rev/s
The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.
![L_{out} = L_{in}](https://tex.z-dn.net/?f=L_%7Bout%7D%20%3D%20L_%7Bin%7D)
![I_{out} \omega_{out} = I_{in} \omega_{in}\\\\\omega_{out} = \frac{ I_{in} \omega_{in} }{I_{out} } \\\\\omega_{out} = \frac{0.9*4}{3.1} \\\\\omega_{out} = 1.161 \ rev/s](https://tex.z-dn.net/?f=I_%7Bout%7D%20%5Comega_%7Bout%7D%20%3D%20I_%7Bin%7D%20%5Comega_%7Bin%7D%5C%5C%5C%5C%5Comega_%7Bout%7D%20%20%3D%20%5Cfrac%7B%20I_%7Bin%7D%20%5Comega_%7Bin%7D%20%7D%7BI_%7Bout%7D%20%7D%20%5C%5C%5C%5C%5Comega_%7Bout%7D%20%20%3D%20%5Cfrac%7B0.9%2A4%7D%7B3.1%7D%20%5C%5C%5C%5C%5Comega_%7Bout%7D%20%20%3D%201.161%20%5C%20rev%2Fs)
Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s