All categories

3 years ago

What system will break the chewed food down to a form your cell can use

1 answer:

4 0

If I'm not mistaken it should be the digestive system due to the fact that our mouths and stomachs break down food and our intestines absorb any water and nutrients

You might be interested in

Match the half life and time information to the percentage of radioactive isotope left.

Kazeer [188]

Answer:

Explanation:

can i have brainliest pls im new

8 0

2 years ago

Anyone know how to solve this? 4+9×2÷3−1

Nutka1998 [239]

You want to use PEMDAS to solve this equation.

4+9*2/3-1

4+18/6-1

4+6-1

10-1

Your answer is 9

Thanks -John

If you have anymore question just ask! :)

5 0

2 years ago

Rex makes himself a bowl of cereal. He puts 28 g of cereal in a bowl and then adds 245 g of milk. Before eating the cereal, he r

Galina-37 [17]

Answer:

A.)

Explanation:

Because you didn't add anything or take anything away.

5 0

2 years ago

A 5kg object moving horizontally at 3m/s collides with a stationary 3kg object. After the collision, the 5kg object is deflected

gavmur [86]

Answer:

The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

Explanation:

Given that,

Mass of object = 5 kg

Speed = 3 m/s

Mass of stationary object = 3 kg

Moving object deflected = 30°

Stationary object deflected = 31°

We need to calculate the velocity of each ball after collision

Using conservation of momentum

Along x-axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\cos\theta+m_{2}v_{2}\cos\theta$

Put the value into the fomrula

$5\times3+0=5\times v_{1}\cos30+3\times v_{2}\cos45$

$15=5v_{1}\times\dfrac{\sqrt{3}}{2}+3v_{2}\times\dfrac{1}{\sqrt{2}}$

$15=\dfrac{5\sqrt{3}}{2}v_{1}+\dfrac{3}{\sqrt{2}}v_{2}$ ....(I)

Along y -axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}\sin\theta+m_{2}v_{2}\sin\theta$

Put the value into the formula

$0+0=5\times v_{1}\sin30-3\times v_{2}\sin45$

$\dfrac{5}{2}v_{1}-\dfrac{3}{\sqrt{2}}v_{2}=0$ ...(II)

From equation (I) and (II)

$v_{1}=\dfrac{15\times2}{5\sqrt{3}+5}$

$v_{1}=2.19\ m/s$

Put the value of v₁ in equation (I)

$\dfrac{5}{2}\times2.19-\dfrac{3}{\sqrt{2}}v_{2}=0$

$v_{2}=\dfrac{5.475\times\sqrt{2}}{3}$

$v_{2}=2.58\ m/s$

Hence, The velocity of each ball after the collision are 2.19 m/s and 2.58 m/s.

3 0

2 years ago

Which Quantity is a vector Quantity? A] Acceleration B] Mass C] Speed D] Volume

vlabodo [156]

Well I am positive that the answer is B

4 0

2 years ago

Read 2 more answers