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3 years ago

How did Earth's atmosphere become oxygen-rich?

Physics

1 answer:

tatyana61 [14]3 years ago

5 0

Trees is the answer

You might be interested in

Many people confuse the terms energy, work, and power. Explain how they are alike and how they are different

nikitadnepr [17]

Work, is the speed at which energy transfers.
Energy, is the capacity to do work.
Power, is the rate at which work is done.

3 0

3 years ago

An arrow of mass 0.5kg so it has 25J of kinetic energy; find the speed of the arrow v = m/s

miss Akunina [59]

Answer:

10 m/s

Explanation:

Use the kinetic energy formula:

KE=(1/2)mv^2

I always remember it as Kevin is half-mad, and very square.

25J = (1/2)*0.5kg*(v^2)

50J = 0.5kg*(v^2)

100J = v^2

v = 10 m/s

Check it:

KE = (1/2)*0.5*(10^2)

KE = 25J

yep, it's right!

4 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

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

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

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

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago

A constant net force that has a magnitude of 135.5 N is exerted on a 26.7 kg object that is initially not moving. a. Draw a free

jek_recluse [69]

Explanation:

Given that,

Net force = 135.5 N

Mass = 26.7 kg

(a). We need to draw a free body diagram

A force is exerted on a object.

(b). We need to calculate the acceleration

Using formula of acceleration

$a =\dfrac{F}{m}$

$a=\dfrac{135.5}{26.7}$

$a=5.07\ m/s^2$

(c). We need to draw the sketch a position vs. time graph for the object.

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times5.07\times t^2$

$s=2.535 t^2$

(d). We need to draw the sketch a velocity vs. time graph for the object.

Using equation of motion

$v = u+at$

Put the value into the formula

$v=0+5.07t$

$v =5.07 t$

(e). We need to calculate the distance travel in 8.82 s

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times5.07\times(8.82)^2$

$s=197.20\ m$

Hence, This is the required solution.

8 0

3 years ago

A piece of gum is stuck to the outer edge of a horizontal turntable, which is revolving at a constant speed. The shadow of the g

cestrela7 [59]

Answer:

angular frequency of the table must be same as the frequency of the projection of the gum on the wall

Explanation:

Since we know that the projection on the wall is the vertical component of the position of the gum on the rotating table

So here we will say

$y = R sin\theta$

so the angle made by the radius vector depends on the angular frequency of the disc by which it is rotating

So we can say

$\theta = \omega t$

so here we can say

$y = R sin(\omega t)$

so here we can say that

angular frequency of the table must be same as the frequency of the projection of the gum on the wall

7 0

4 years ago