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3 years ago

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Pure sodium metal placed in water will spark and ignite, as well as form bubbles and gas. What are the signs that this is a chem

ical reaction? Check all that apply.
Heat is produced.

Light is produced.

Heat is absorbed.

Bubbles appear.

A precipitate forms.

The color changes.

2 answers:

ad-work [718]3 years ago

7 0

<h3><u>Answer;</u></h3>

Heat is produced
Light is produced
Bubbles appear

<h3><u>Explanation;</u></h3>

Chemical reactions involve the chemical interaction of two or more chemical substances, result in a new substance being formed, and are usually irreversible.
The signs of chemical reactions include gas formation, energy release in the form of light or flame, heat absorption, precipitate formation, and color change.

Aleonysh [2.5K]3 years ago

6 0

Answer:

Heat is produced

Light is produced

Bubbles appear

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Explanation:

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two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y

svet-max [94.6K]

Answer:

Explanation:

Given

initial velocity component of engines is

$v_0_x=6380 m/s$

$v_0_y=6770 m/s$

time period of engine running=763 s

Displacement in $x=4.50\times 10^6$

$y=7.27\times 10^6$

Using $s=ut+\frac{at^2}{2}$ in x and y direction

$x=v_0_x\times t+\frac{at^2}{2}$

$4.50\times 10^6=6380\times 763+\frac{a\times 763^2}{2}$

$4.50\times 10^6-4.86\times 10^6=\frac{a\times 763^2}{2}$

$a=-1.23 m/s^2$

In y direction

$y=v_0_y\times t+\frac{a't^2}{2}$

$7.27\times 10^6=6770\times 763+\frac{a\times 763^2}{2}$

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A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

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Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

$F-f=ma$

Put the value into the formula

$340-f=60\times2.0$

$f=340-60\times2.0$

$f=220\ N$

We need to calculate the coefficient of friction

Using formula of friction force

$f= \mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{220}{60\times9.8}$

$\mu =0.37$

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

$F = ma$

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

$a=\dfrac{220}{100}$

$a=2.2\ m/s^2$

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

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4 years ago