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3 years ago

15

A scientist is examining an unknown solid. Which procedure would most likely help determine a chemical property of the substance

?

2 answers:

Helga [31]3 years ago

6 0

Answer:

exposing it to a flame to see if it catches on fire

Explanation:

edge 2020

Neporo4naja [7]3 years ago

4 0

<span>procedure would most likely help determine a chemical property of the substance is : exposing it to a flame to see if it catches on fire Chemical property is the characteristic that a substance has that differntiate it with another substance. The most common charatcteristics that most scientists wanted to know are : - It's flamability - It's radioactivity - Its toxicity By throwing the object into fire, we will easily find out these 3 characteristics</span>

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What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

Wittaler [7]

Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

6 0

3 years ago

An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

Marina86 [1]

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

4 0

3 years ago

HELP ASAP! GIVING BRAINLIEST!

Blababa [14]

Answer:

1. Emma standing on top of mountain

Since she is at the rest position and at some height from the ground so here energy is due to gravitational potential energy

So we have

gravitational potential energy

$U = mgH$

2. Emma jumping down from mountain top

Due to free fall Emma will start moving with some speed in downwards direction so here we have

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

motion energy

3. tension in rope at Emma’s lowest position

Due to stretch in the rope here position come to the lowest end and speed comes to zero so whole energy is converted into elastic potential energy

$U = \frac{1}{2}kx^2$

elastic potential energy

4. Emma bouncing back

Due to bouncing back she will again have its kinetic energy with some speed upwards

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

motion energy

8 0

3 years ago

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Where is the electric field zero? a. region 2, 0.46 m from the +7 µC charge b. region 2, 0.46 m from the +5 µC charge c. the fie

Ratling [72]

Answer:

the field is not zero anywhere on the x axis (except at infinity)

Explanation:

From the Coulomb's law we have electric field intensity as:

$E=\frac{1}{4\pi.\epsilon_o} \frac{q}{r^2}$

where:

$\epsilon_o=$ permittivity of free space

$q=$ charge due to which the field is generated.

$r=$ distance from the charge

So, from the above relation:

Electric field due to a charge can only be zero at infinity.

3 0

4 years ago

Semi-conductors are materials that

mixas84 [53]

You're right, it is C

6 0

3 years ago

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