An alcohol functional group is called a hydroxyl group.
It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left
Answer:
The shadow will appear dark
Explanation:
Your shadow is caused by your body blocking incident light rays from reach the ground under the shadow area. The absence of light means there will be no light reflected from the ground making it appear dark.
Answer:
Vacculoes, vesicles
Explanation:
Organelles and other structures within the cells have vesicles collectively referred to as the endo-membrane system. The Lysosomes pack up the substances, and the vesicles go through the porsous membrame and secrete the substances.
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force



Distance between planet and star

Gravitational force is

Applying the new distance,


Replacing with the previous force,

Replacing our values


Therefore the magnitude of the force on the star due to the planet is 