a. of water a. of bread a. of soap a. of juice a. of packet a. of sand option_piece packet grain bottle cake slice bale tank pane note tube bar can sack pile sheaf​

Answers

Answer 1
Answer:

Answer:

is it a poem or recipe :(

don't know


Related Questions

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What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.The pressure everywhere increases by the same amount.The pressure everywhere decreases to conserve total pressure.
A ship is traveling at 154 m/s and accelerates at a rate of 1.80 m/s^2 for 1 minute. What will its speed be after that minute? Calculate the answer in both meters per second and kilometers per hour.
The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?
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Maggots feed on dead and decaying organisms for energy. What are maggots?autotrophs
producers
decomposers
heterotrophs

Answers

Answer:

Explanation:

Decomposers is the correct answer

Answer:

Decomposers is the right answer

Explanation:

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?

Answers

Answer:

a.) Speed V = 29.3 m/s

b.) K.E = 1931.6 J

Explanation: Please find the attached files for the solution

Final answer:

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.

Explanation:

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:

brainly.com/question/13897993

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PLEASE HELP!!! Sophia says that if we add two electrons to oxygen in order to fill its valence shell, it’s expected charge would be +2. Is she correct? If not, explain the error in her thinking.

Answers

The error in her thinking is that oxygen has has six electrons and a negative charge is acquired by nitrogen when it gains two electrons.

Oxygen is a member of group 16. The elements in group 16 has six valence electrons. This means that they need an extra two electrons to complete their octet.

If an atom gains two electrons, it will have a charge of -2 and not +2, a positive charge means that the atom lost electrons. Nonmetals like oxygen do not loose electrons rather they gain electrons.

Learn more: brainly.com/question/14156701

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

Answers

Answer:

B

Explanation:

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.

Answers

Answer:

av=0.333m/s, U=3.3466J

b.

v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s

Explanation:

a. let m_A be the mass of block A, andm_B=10.0kg be the mass of block B. The initial velocity of A,\rightarrow v_A_1=2.0m/s

-The initial momentum =Final momentum since there's no external net forces.

pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

v_A_1-v_B_1=v_(B2)-v_(A2)

-Applying the conservation of momentum. The blocks have the same velocity after collision:

v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s

#Total Mechanical energy before and after the elastic collision is equal:

K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)

2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

Answers

Answer:

15 m/s

Explanation:

We know that v = f * d where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s