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2 years ago

9

After food is consumed, chemical nutrients are absorbed into the blood. The blood then carries these nutrients to the body's cel

ls where the nutrients enter mitochondria. Which reaction must occur next for the cell to carry on other activities?

1 answer:

guapka [62]2 years ago

3 0

Answer:b

Explanation:b

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What is the definition of electrical power?

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An electric power measure the rate of electrical energy transfer by an electric circuit per unit of time.

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2 years ago

Which weather condition commonly occurs along a cold front

aleksklad [387]

Precipitation is your answer. according to my science textbook, in a cold front, warm air and cold air interact and cause precipitation.

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3 years ago

Read 2 more answers

A ball is thrown vertically upward from the ground. Its distance in feet from the ground in t seconds is s equals negative 16 t

marishachu [46]

Answer:

t₁ = 3 s

Explanation:

In this exercise, the vertical displacement equation is not given

y = 240 t + 16 t²

Where y is the displacement, 240 is the initial velocity and 16 is half the value of the acceleration

Let's replace

864 = 240 t + 16 t²

Let's solve the second degree equation

16 t² + 240 t - 864 = 0

Let's divide by 16

t² + 15 t - 54 = 0

The solution of this equation is

t = [-15 ± √(15 2 - 4 1 (-54)) ] / 2 1

t = [-15 ±√(225 +216)] / 2

t = [-15 + - 21] / 2

We have two solutions.

t₁ = [-15 +21] / 2

t₁ = 3 s

t₂ = -18 s

Since time cannot have negative values, the correct t₁ = 3s

4 0

3 years ago

Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

Zigmanuir [339]

Answer:

Part a)

$I = 1.5 kg m^2$

Part b)

$I = 0.75 kg m^2$

Part c)

$I = 1.5 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about an axis passing through B and C is given as

$I = mL^2 + mL^2 + m(0) + m(0)$

$I = 2mL^2$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

Part b)

Moment of inertia of the system about an axis passing through A and C is given as

$I = m(0^2) + m(\frac{L}{\sqrt2})^2 + m(0) + m(\frac{L}{\sqrt2})^2$

$I = 2m\frac{L^2}{2}$

$I = (3 kg)(0.50^2)$

$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

$I = m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2 + m(\frac{L}{\sqrt2})^2$

$I = 4m\frac{L^2}{2}$

$I = 2(3 kg)(0.50^2)$

$I = 1.5 kg m^2$

8 0

3 years ago

The young tree was bent and has been brought into a vertical position by the three guy cables. If tension at AB = 0, AC = 10 lb,

KATRIN_1 [288]

Answer:

The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.

C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'

Explanation:

See attached picture.

3 0

3 years ago