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4 years ago

How does a rubber rod become negatively charged through friction?

Physics

1 answer:

mojhsa [17]4 years ago

3 0

Answer:

c. It is rubbed with another object, and electrons move onto the rod.

Explanation:

A rubber rod is negatively charged through friction when gains electrons, this occurs when electrons are transferred from another material by simple contact between them. When these two materials are rubbed, the difference in electronic affinity causes one of the materials to be positively charged and the other negatively charged.

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A living thing that feeds on Another living thing and may kill it eventually is called

Murljashka [212]

Answer:

(D) parasite........................

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2 years ago

A ray of monochromatic light ( f = 5.09 × 10^14 Hz) passes from air into Lucite at an angle

harina [27]

Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:

$\frac{sen\O_{2}}{sen\O_{1}} = \frac{n_{2}}{n_{1}} \\ sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}}$

Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:

$sen\O_{2}= \frac{n_{2}*sen\O_{1}}{n_{1}} \\ sen\O_{2}= \frac{1.5* \frac{1}{2} }{1} \\ sen\O_{2}=0.75$

Using the arcsin properties, we get:

$sen\O_{2}=0.75 \\ arcsin(0.75)=\O_{2} \\ \boxed {\O_{2}=48.59^o}$

Obs: Approximate results, and the drawing is attached

If you notice any mistake in my english, let me know, because i am not native.

6 0

3 years ago

if an object is thrown directly upwards remains in the air for 5.6s before it returns to its original position, how long did it

Mila [183]

Okay so an object gets thrown upwards so the half way point of the trip would be the maximum height (before it starts coming back down). if the object stays in the air for a total of 5.6s, then that is the time it takes to go up and come back down. to find the time to maximum height, half the time of the whole trip

5.6s/2 = 2.8s

6 0

4 years ago

Read 2 more answers

Which tool would you use to measure the height of a cube of metal

SCORPION-xisa [38]

It depends on the size of the cube, but you could use a ruler for almost any size, or if it is small enough, you could use a caliper to get a very precise measurement.

3 0

3 years ago

Read 2 more answers

The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o

matrenka [14]

Answers:

(a) $0.0073kg$

(b) Volume gold: $3.79(10)^{-7}m^{3}$ , Volume cupper: $7.6(10)^{-8}m^{3}$

(c) $17633.554kg/m^{3}$

Explanation:

We are told the total mass $M$ of the coin, which is an alloy of gold and copper is:

$M=m_{gold}+m_{copper}=7.988g=0.007988kg$ (1)

Where $m_{gold}$ is the mass of gold and $m_{copper}$ is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats $K$ and mass is:

$K=24\frac{m_{gold}}{M}$ (2)

Finding ${m_{gold}$ :

$m_{gold}=\frac{22}{24}M$ (3)

$m_{gold}=\frac{22}{24}(0.007988kg)$ (4)

$m_{gold}=0.0073kg$ (5) This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density $\rho$ of an object is given by:

$\rho=\frac{mass}{volume}$

If we want to find the volume, this expression changes to: $volume=\frac{mass}{\rho}$

For gold, its volume $V_{gold}$ will be a relation between its mass $m_{gold}$ (found in (5)) and its density $\rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}$ :

$V_{gold}=\frac{m_{gold}}{\rho_{gold}}$ (6)

$V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}$ (7)

$V_{gold}=3.79(10)^{-7}m^{3}$ (8) Volume of gold in the coin

For copper, its volume $V_{copper}$ will be a relation between its mass $m_{copper}$ and its density $\rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}$ :

$V_{copper}=\frac{m_{copper}}{\rho_{copper}}$ (9)

The mass of copper can be found by isolating $m_{copper}$ from (1):

$M=m_{gold}+m_{copper}$

$m_{copper}=M-m_{gold}$ (10)

Knowing the mass of gold found in (5):

$m_{copper}=0.007988kg-0.0073kg=0.000688kg$ (11)

Now we can find the volume of copper:

$V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}$ (12)

$V_{copper}=7.6(10)^{-8}m^{3}$ (13) Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density $\rho_{coin$ will be a relation between its total mass $M$ and its total volume $V$ :

$\rho_{coin}=\frac{M}{V}$ (14)

Knowing the total volume of the coin is:

$V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3}$ (15)

$\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}}$ (16)

Finally:

$\rho_{coin}=17633.554kg/m^{3}}$ (17) This is the total density of the British sovereign coin

6 0

3 years ago