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3 years ago

List an example of each of the four classes polymers that living things make and use

Physics

1 answer:

slamgirl [31]3 years ago

6 0

The four classes of polymers are:

1. Nucleic acids. Examples are DNA and RNA

2. Protein. Examples are enzymes and hemoglobin

3. Carbohydrates. Examples as starch and glycogen

4. Lipids. Examples are triglycerides and phospholipids

The building blocks of nucleic acids are called bases and there are four types known as Guanine, Adenine, Thymine and Cytosine.

The building blocks of carbohydrates are glucose molecules.

The building blocks of protein are amino acids.

The building blocks of lipids are a combination of fatty acids and glycerol.

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What type of machine are wire cutter pliers?

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It a compound machine

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Compared to its weight on Earth, a 10-kg object on the moon will weigh Question 9 options: less. the same amount. more.

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it will weigh less on the moon.

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What is the momentum of a 2000 kg mass moving at 3 meters per second

podryga [215]

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 2000 × 3

We have the final answer as

Hope this helps you

3 0

3 years ago

PLEASE HELP ASAP

morpeh [17]

A is the correct answer

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3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

We can use the equation of the horizontal position here.

<u>First stone</u>

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

<u>Second stone</u>

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago