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3 years ago

When water vapor is cooled it three forms into water droplets why does this represent a conservation in mass

1 answer:

8 0

In the conservation of mass, mass is never created or destroyed in chemical reactions in the same way water is not created or destroyed it is only transferred from one form to another and its mass is always conserved.

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WHAT STRUCTURE DO ALL OF BORON, CARBON, NITROGEN, OXYGEN, AND FLOURINE HAVE IN COMMON

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3 years ago

What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 N at an angle of 35Â° above the

Olegator [25]

The acceleration of the wagon along the ground is 3.6 m/s².

To solve the problem above, we need to use the formula of acceleration as related to force and mass.

Acceleration: This can be defined as the rate of change of velocity.

⇒ Formula:

Fcos∅ = ma................. Equation 1

⇒ Where:

F = Force
∅ = angle above the horizontal
m = mass of the wagon
a = acceleration of the wagon

⇒ make a the subject of equation 1

a = Fcos∅/m..................... Equation 2

From the question,

⇒ Given:

F = 44 N
∅ = 35°
m = 10 kg

⇒ Substitute these values into equation 2

a = 44(cos35°)/10
a = 44(0.8191)/10
a = 3.6 m/s²

Hence, The acceleration of the wagon along the ground is 3.6 m/s²

Learn more about acceleration here: brainly.com/question/9408577

3 0

3 years ago

What factors in Earth's relationship to the Sun make life on Earth possible?

Mila [183]

The answer is B.

The planet cannot be too hot or too cold it has to be the right distance from its sun to maintain life.

5 0

3 years ago

Read 2 more answers

A rolling ball has an initial velocity of 1.6 meters per second. if the ball has a constant acceleration of 0.33 meters per seco

lukranit [14]

We have:

Initial velocity (u) = 1.6 m/s
Constant acceleration (a) = 0.33 m/s²
Time (t) = 3.6 sec

There are five constant acceleration equations that would help us to find the velocity:

$v=u+at$
$s=ut+ \frac{1}{2}at^2$
$s= \frac{1}{2}(u+v)t$
$v^2=u^2+2as$
$s=vt- \frac{1}{2}at^2$

Since we have $u, a, t$ and we want $v$
We will use the first formula $v=u+at$

$v=1.6+(0.33)(3.6)$
$v= 2.788$ m/s

7 0

3 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago