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4 years ago

10

Balance this chemical equation Fe(s) + 02(g) --- Fe0(s)

1 answer:

Softa [21]4 years ago

6 0

2Fe(s) + O2 -> 2FeO(s)

<span>2 'Fe' atoms on both sides </span>
<span>2 'O' atoms on both sides</span>

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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago

For this situation to demonstrate a balanced force, how much force must Omar apply? (In the picture, Sam is pushing a box to the

11111nata11111 [884]

The answers 10N as it equals it out

3 0

3 years ago

Read 2 more answers

I am giving a reward of 17 points do you want more rewards?

stiks02 [169]

Answer: I do not mind if I have more rewards or not.

Explanation: I think helping others is the best reward

4 0

1 year ago

Suppose a two-level system is in a bath with temperature 247 K. The energy difference between the two states is 1.1 × 10-21 J. W

Alik [6]

Answer:

The probability of higher energy state is 0.4200.

Explanation:

Given that,

Temperature = 247 K

Energy difference between two states $E_{2}-E_{1}=1.1\times10^{-21}\ J$

We need to calculate the probability of higher energy state

Probability of $E_{1}= e^{-\beta E_{1}}$

Probability of $E_{2}= e^{-\beta E_{2}}$

The total probability is

$e^{-\beta E_{1}}+e^{-\beta E_{1}}=1$

Here, E₁ = lower energy state

E₂ = higher energy state

Put the value of E₁ in to the formula

$e^{-\beta(E_{2}-1.1\times10^{-21})}+e^{-\beta E_{1}}=1$

$e^{-\beta E_{2}}(e^{\beta 1.1\times10^{-21}}+1)=1$

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{KT}}}$

Here, $\beta=\dfrac{1}{KT}$

Put the value into the formula

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{1.380\times10^{-23}\times247}}}$

$e^{\beta E_{2}}=0.4200$

Hence, The probability of higher energy state is 0.4200.

4 0

3 years ago

A 1.0 kg ball is thrown into the air with an initial veocity of 30. m/s. how much kinetic energy does the ball have?

rosijanka [135]

1/2 mv2 so you get 450j

8 0

4 years ago