12.) Active transport because the cell must use energy to move large particles across the membrane.
13.) Photosynthesis takes place in plant leaves containing the chlorophyll pigment. Cellular respiration takes place in the cytoplasm and mitochondria of the cell. ... Cellular respiration uses glucose molecules and oxygen to produce ATP molecules and carbon dioxide as the by-product.
14.) In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into two main stages: interphase and the mitotic (M) phase (including mitosis and cytokinesis). During interphase, the cell grows, accumulating nutrients needed for mitosis, and undergoes DNA replication preparing it for cell division.
Answer: c
Explanation:
C Air is a compound of two or more components that keep their own identifying properties, while water is composed of mixtures that combine to form a compound.
AnswerA tell me what u got sorry if its worng
Explanation:
Explanation:
Given that,
Mass = 0.254 kg
Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]
Force = 0.5 N
y = 0.628
We need to calculate the A and d
Using formula of A and d
.....(I)
....(II)
Put the value of
in equation (I) and (II)


From equation (II)


Put the value of
in equation (I) and (II)


From equation (II)


Put the value of
in equation (I) and (II)


From equation (II)


Put the value of
in equation (I) and (II)


From equation (II)


Hence, This is the required solution.