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4 years ago

What would lack of calcium do to bones

Physics

2 answers:

Sonja [21]4 years ago

5 0

They compromise the density of the bones causing them to be brittle and easy to break hope it helps

timurjin [86]4 years ago

4 0

It would weaken the bones and thy would not be as strong or tough

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What type of cells don't have cell walls.

iren [92.7K]

Answer:

Animal Cells

Explanation:

they dont have cell walls because they dont need them. Wall cells are found in plants, which animals dont need.

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3 years ago

Read 2 more answers

Would an increase in pressure favor the formation of ozone or of oxygen?

max2010maxim [7]

<h3><u>Answer;</u></h3>

<u>An increase in pressure favors the formation of ozone </u>

<h3><u>Explanation;</u></h3>

Ozone, O3, decomposes to molecular oxygen in the stratosphere according to the reaction

2O3(g) ⇆ 3O2 (g).

There are more moles of product gas than moles of reactant gas. An increase in total pressure increases the partial pressure of each gas, shifting the equilibrium towards the reactants.
Therefore; an increase in pressure favors backward reactions towards the formation of ozone.

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3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as