Help will mark brain list

PLEASEEE HELP ME WITH THIS ALSO. I DONT WANT TO FAIL. You push a merry-go-round on which Kim and Katie are riding. Kim weighs 45

Serjik [45]

Answer:

The body weight

Explanation:

5 0

3 years ago

The waste product of photosynthesis is:

zalisa [80]

Answer:

The waste product of photosynthesis is oxygen

6 0

2 years ago

Please Answer These Questions I Really Need Help Please answer these questions

wel

1. Onion cells have thick rectangular walls.
2. Both are tropospheres, the lobster and fungi have the same outer shell.
3. A bat is a mammal--it has fur, lactates and is warm-blooded. Mammals are vertebrates. Vertebrates have a dorsal nerve cord protected by bony or cartilaginous vertebrae. Arthropods are invertebrates--they do not have vertebrae. They are a specific kind of invertebrate with a jointed exoskeleton.
4, 5, 6, 7 - Sorry don't know these answers to these questions

8 0

2 years ago

A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo

Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

$E_{1}=E_{pot}+E_{kin}+E_{elas}\\$

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

$E_{2}=E_{kin}+E_{pot}$

Now we can use the first statement to get the first equation:

$E_{1}+W_{1-2}=E_{2}$

where:

W₁₋₂ = work from the state 1 to 2.

$E_{k}=\frac{1}{2} *m*v^{2} \\$

$E_{pot}=m*g*h$

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

$70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5$

$58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]$

4 0

3 years ago

Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit

cestrela7 [59]

Answer:

$v=1667.9km/h$

$a_{cp}=436.6km/h^2$

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth $C=2\pi R$ , where $R=6371km$ is the radius of the Earth.

We have then:

$v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h$

And then we use the centripetal acceleration formula:

$a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2$

6 0

2 years ago