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3 years ago

Which equation represents an equilibrium system??

Physics

1 answer:

MrRissso [65]3 years ago

7 0

A. 2 S O 2, gas, plus O 2, gas, in equilibrium with 2 S O 3, gas.

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A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.

EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

m is mass of the projectile
v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0

1 year ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to <u>Newton's 2nd law of motion:</u>

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

7 0

2 years ago

An observer standing near a window 5 m high observe that an object falling downwards is passing across the window in 0.5 s. Find

Valentin [98]

Answer:

Explanation:

An object in free fall, NOT experiencing parabolic motion, has an equation of

$h(t)=\frac{1}{2} gt^2+h_0$ which says:

The height of an object with respect to time in seconds is equal to the pull of gravity times time-squared plus the height from which it was dropped. Normally we use -9.8 for gravity but you said to use 10, so be it.

For us, h(t) is 5 because we are looking for the height of the window when the object is 5 m off the ground at .5 seconds;

g = 10 m/s/s, and

t = .5sec

$5=\frac{1}{2}(-10)(.5)^2$ +h and

5 = -5(.5)² + h and

5 = -5(.25) + h and

5 = -1.25 + h so

h = 6.25

That's how high the window is above the ground.

8 0

2 years ago

Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,

Nutka1998 [239]

Good morning.

We have:

$\mathsf{\overset{\to}{a} = 29\overset{\to}{j}}$

Where j is the unitary vector in the direction of the y-axis.

We have that

$\mathsf{\overset{\to}{a}+\overset{\to}{b} = -18\overset{\to}{j}}$

We add the vector -a to both sides:

$\mathsf{\overset{\to}{b} = -18\overset{\to}{j} -\overset{\to}{a} = -18\overset{\to}{j} -29\overset{\to}{j}}\\ \\ \mathsf{\overset{\to}{b}=-47\overset{\to}{j}}$

Therefore, the magnitude of b is 47 units.

5 0

3 years ago

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