All categories

4 years ago

A boxer is competing in a match. Which describes the roles of endocrine and neural controls as he boxes?

Biology

2 answers:

Alisiya [41]4 years ago

8 0

Answer:

Option A

Explanation:

In case of match when a boxer is playing a match. Then there is a role of both nervous system and endocrine system in the body.

The player is feeling excited and at the same time he is also feeling pain as he is boxer he might have got some punches from the opponent player.

Here, the nervous system transmits pain and the endocrine system of the body will release hormones as a response of fight or flight conditions.

Afina-wow [57]4 years ago

4 0

I think the answer your looking 4 is B

HOPE THIS HELPS
#Optimist

You might be interested in

In which season did the rabbit population increase the least

Liula [17]

Answer:

Winter

Explanation:

6 0

3 years ago

Read 2 more answers

Which of the following is likely to happen in a deforested area?

zheka24 [161]

Increase in soil erosion. hop it helps.

8 0

3 years ago

Read 2 more answers

Explain how glaciers can change the landscape and physical characteristics of continents.

kompoz [17]

A glacier's weight, combined with its gradual movement, can drastically reshape the landscape over hundreds or even thousands of years. The ice erodes the land surface and carries the broken rocks and soil debris far from their original places, resulting in some interesting glacial landforms.

By their shear weigh,t glaciers grind and sculpt the land. As a glacier travels it picks up large boulders and smaller debris and transports them down hill.

3 0

3 years ago

Read 2 more answers

Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

$C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}$

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

Best regards.

7 0

3 years ago

How does the vacuole (10) in this animal cell differ from the vacuole in a plant cell

dexar [7]

The vacuole in an animal cell is small and temporary
WHILE
The vacuole in a plant cell is large and permanent

3 0

4 years ago