Maggots feed on dead and decaying organisms for energy. What are maggots?autotrophs producers decomposers heterotrophs
Decomposers is the correct answer
Decomposers is the right answer
Maggots are decomposers because they eat the dead bodys for energy
I don't know if the thing I wrote it truse so ya
A bicycle racer is going downhill at 11.0 m/s when, to his horror, one of his 2.25 kg wheels comes off when he is 75.0 m above the foot of the hill. We can model the wheel as a thin-walled cylinder 85.0 cm in diameter and neglect the small mass of the spokes. (a) How fast is the wheel moving when it reaches the bottom of hill if it rolled without slipping all the way down? (b) How much total kinetic energy does it have when it reaches bottom of hill?
a.) Speed V = 29.3 m/s
b.) K.E = 1931.6 J
Explanation: Please find the attached files for the solution
The wheel's speed at the bottom of the hill can be found through the conservation of energy equation considering both translational and rotational kinetic energy, while the total kinetic energy at the bottom of the hill is a sum of translational and rotational kinetic energy.
These two questions address the physics concepts of conservation of energy, kinetic energy, and rotational motion. To answer the first question, (a) How fast is the wheel moving when it reaches the bottom of the hill if it rolled without slipping all the way down?, we need to consider the potential energy the wheel has at the top of the hill is completely converted into kinetic energy at the bottom. This includes both translational and rotational kinetic energy. Solving for the final velocity, vf, which would be the speed of the wheel, we get vf = sqrt((2*g*h)/(1+I/(m*r^2))), where g is the acceleration due to gravity, h is the height of the hill, I is the moment of inertia of the wheel, m is the mass of the wheel, and r is the radius of the wheel.
For the second question, (b) How much total kinetic energy does it have when it reaches bottom of the hill?, we use the formula for total kinetic energy at the bottom of the hill, K= 0.5*m*v^2+0.5*I*(v/r)^2. Substituting the value of v found in the first part we find the kinetic energy which we can use the formula provided in the reference information.
Learn more about Conservation of Energy and Rotational Motion here:
According to American Heart Association, your target heart rate is__________. Group of answer choices
A .all of the above
B. 85% - 95% of your max heart rate
C. 20%-30% of your max heart rate
D. 50%-85% of your max heart rate
A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal
Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.
a. let be the mass of block A, and be the mass of block B. The initial velocity of A,
-The initial momentum =Final momentum since there's no external net forces.
Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):
-Applying the conservation of momentum. The blocks have the same velocity after collision:
#Total Mechanical energy before and after the elastic collision is equal:
Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s
b. Taking the end collision:
From a above,
We plug these values in the equation:
What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?
We know that where f = frequency & d = wavelength .