Look up on google Transcript of Similarities and Differences Among Living Organisms. Controls the transport of materials into and out of the cell. ... For example, fish, amphibians, reptiles, birds, and mammals reproduce sexually Humans use the same kind of genetic information as nearly every living organism - DNA!
100/2.5 is 40.
40 is the energy that is being produced
Answer:
d = 2.54 [m]
Explanation:
Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

where:
W = work [J]
Ek1 = kinetic energy at initial state [J]
Ek2 = kinetic energy at the final state = 0.
We must remember that kinetic energy can be calculated by means of the following expression.
![\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D-W%3D0%5C%5CW%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A4%2A%285%29%5E%7B2%7D%5C%5CW%3D%2050%20%5BJ%5D)
We know that work is defined as the product of force by distance.

where:
F = force [N]
d = distance [m]
But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.
![f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]](https://tex.z-dn.net/?f=f%3Dm%2Ag%2A0.5%5C%5Cf%20%3D%204%2A9.81%2A0.5%5C%5Cf%20%3D%2019.62%20%5BN%5D)
Now solving the equation for the work.
![d=W/F\\d = 50/19.62\\d = 2.54[m]](https://tex.z-dn.net/?f=d%3DW%2FF%5C%5Cd%20%3D%2050%2F19.62%5C%5Cd%20%3D%202.54%5Bm%5D)
Answer:

Explanation:
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e.
and 
The wavelength of Hi line of the Balmer series is given by :




So, the wavelength for this line is 550 nm. Hence, this is the required solution.
Answer:
The unknown substance is Aluminum.
Explanation:
We'll begin by calculating the change in the temperature of substance. This can be obtained as follow:
Initial temperature (T₁) = 25 ⁰C
Final temperature (T₂) = 100 ⁰C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 100 – 25
ΔT = 75 ⁰C
Finally, we shall determine the specific heat capacity of the substance. This can be obtained as follow:
Change in temperature (ΔT) = 75 ⁰C
Mass of the substance (M) = 135 g
Heat (Q) gained = 9133 J
Specific heat capacity (C) of substance =?
Q = MCΔT
9133 = 135 × C × 75
9133 = 10125 × C
Divide both side by 10125
C = 9133 / 10125
C = 0.902 J/gºC
Thus, the specific heat capacity of substance is 0.902 J/gºC
Comparing the specific heat capacity (i.e 0.902 J/gºC) of substance to those given in the table above, we can see clearly that the unknown substance is aluminum.