Answer: reflected, clouds, greenhouse effect.
Explanation:
Sunlight can be absorbed, reflected , or scattered before it reaches Earth’s surface. About 30 percent of sunlight hits Earth directly, and 22 percent is filtered through clouds . Dust particles scatter short wavelengths, which causes the sky to appear blue. Earth radiates heat into the atmosphere, which traps the heat in gases, causing the greenhouse effect.
Answer:
Positron emission
Explanation:
Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.
A positron is usually ejected in the process together with an anti-neutrino to balance the spins.
1) Calculate the number of moles of Cu SO4 . 5H20 by dividing the specified mass by the molar mass.
2) The ratio of production given by the equation is 1 mol of Cu SO4 . 5 H2O to 1 mol of Cu SO4=> 1:1, meaning that the number of moles of Cu SO4 produced is the same number of moles of Cu SO4.5H20 heated.
3) Finally mutiply the number of moles of Cu SO4 by its molar mass and there you have the mass of Cu SO4 produced.
Answer: For transverse waves, the waves move in perpendicular direction to the source of vibration.
For longitudinal waves, the waves move in parallel direction to the source of vibration .
They are similar within the sense that energy is transferred within the kind of waves.
Explanation:
Answer:
Option 2, Half of the active sites are occupied by substrate
Explanation:
Michaelis-Menten expression for enzyme catalysed equation is as follows:
![V_0=\frac{V_{max\ [S]}}{k_M+[S]}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%5C%20%5BS%5D%7D%7D%7Bk_M%2B%5BS%5D%7D)
Here, 
is Michaelis-Menten constant and [S] is substrate concentration.
When [S]=Km
Rearrange the above equation as follows:
![V_0=\frac{V_{max}[S]}{k_M+[S]}\\V_0=\frac{V_{max}[S]}{[S]+[S]}\\V_0=\frac{V_{max}[S]}{2[S\\]}\\V_0=\frac{V_{max}}{2}](https://tex.z-dn.net/?f=V_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7Bk_M%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B%5BS%5D%2B%5BS%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%5BS%5D%7D%7B2%5BS%5C%5C%5D%7D%5C%5CV_0%3D%5Cfrac%7BV_%7Bmax%7D%7D%7B2%7D)
when [S]=Km, the rate of enzyme catalysed reaction becomes half of the maximum rate, that means half of the active sites are occupied by substrate.
Therefore, the correct option is option 2.
