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4 years ago

Plz answer this! I am stumped

Physics

1 answer:

Gwar [14]4 years ago

6 0

MgCl2
Mg = magnesium
Cl = chlorine

Magnesium + chlorine = magnesium chloride.

This is because compounds are always written with the METAL FIRST and the NON METAL SECOND. the non metal ends in - ide when it reacts with a metal.

So ur answer would be magnesium chloride. :)

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Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Oksanka [162]

a) The acceleration of the objects is $3.56 m/s^2$

b) The tension in the string is 280.8 N

Explanation:

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

$m_1 g - T = m_1 a$ (1)

where

$m_1 = 45.0 kg$ is the mass

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

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

$T-m_2 g = m_2 a$ (2)

where

T is the tension in the string

$m_2 = 21.0 kg$ is the mass

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

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

$T=m_1g - m_1 a$

And substituting into (2),

$(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2$

From the previous part of the problem we got an expression for the tension in the string:

$T=m_1g - m_1 a$

Where we have

$m_1 = 45.0 kg$

$g=9.8 m/s^2$

$a=3.56 m/s^2$ is the acceleration, found in part a)

Susbtituting, we find

$T=(45.0)(9.8)-(45.0)(3.56)=280.8 N$

Learn more about forces and acceleration:

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3 years ago

A book weighing 5 N rests on top of a table. 1) A downward force of magnitude 5 N is exerted on the book by the force of

andreev551 [17]

Answer:

1) A downward force of magnitude 5 N is exerted on the book by the force of of gravity

2) An upward force of magnitude 5 N is exerted on the book by the table

Explanation:

First of all, any object near the Earth's surface experiences the forces of gravity, which is also called weight of the object. This force always acts downward.

For the book in the problem, the magnitude of the weight is 5 N.

We also know that the book is at rest: this means that the net force acting on it is zero, and there must be another force balancing the weight, in order to give a zero net force. This other force is the reaction force exerted by the table on the book: the magnitude of this force must be equal to the force of gravity (so, 5 N) and its direction is opposite to the weight, therefore upward.

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3 years ago

1. Two students have volunteered to explore the galaxy in a

LiRa [457]

Answer:a

Explanation:

I guess

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3 years ago

A small car with mass of 0.800 kg travels at a constant speed

Alexandra [31]

Answer:

The equation of equilibrium at the top of the vertical circle is:

\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}

The speed experimented by the car is:

\frac{N}{m}+g=\frac{v^{2}}{R}

v = \sqrt{R\cdot (\frac{N}{m}+g) }

v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}

v\approx 9.302\,\frac{m}{s}

The equation of equilibrium at the bottom of the vertical circle is:

\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}

The normal force on the car when it is at the bottom of the track is:

N=m\cdot (\frac{v^{2}}{R}+g )

N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)

N=21.690\,N

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2 years ago

A rod of length L and electrical resistance R moves through a constant uniform magnetic field ; both the magnetic field and the

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Answer:

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3 years ago