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3 years ago

Which statement about the physical properties of matter is true?

Physics

1 answer:

lisov135 [29]3 years ago

4 0

I believe your answer would be D- Physical properties of matter effect the chemical properties of matter.

Hope this helps. Tell me if I'm right.

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Which of the following represents an upright image? A. -do B. +m C. -m D. +do

Tom [10]

Answer:

B. +m

Explanation:

The magnification of an image is defined as the ratio between the size of the image and of the object:

$m = \frac{y'}{y}$

where we have

y' = size of the image

y = size of the object

There are two possible situations:

- When m is positive, y' has same sign as y: this means that the image image is upright

- When m is negative, y' has opposite sign to y: this means that the image is upside down

Therefore, the correct option representing an upright image is

B. +m

5 0

3 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

3 years ago

Hypothetically speaking, if an object were located at the center of the Earth, the gravitational force on that object due to the

Likurg_2 [28]

Answer:

D. ) The force would be zero newtons

Explanation:

Because

If you are at the center of the earth, gravity is zero because all the mass around you is pulling "up" (every direction there is up!)

So F=mg so if g is zero F is also zero

5 0

3 years ago

Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We

Katen [24]

Answer:

$q=\frac{mg}{E_o}$

Explanation:

Given:

Charge = q

Electric field strength = $E_o$

weight of the droplet = mg

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet, $F=qE_o$

This is balanced by the weight, $mg$

Equating the two:

$qE_o=mg\\\Rightarrow q=\frac{mg}{E_o}$

4 0

3 years ago

A parallel plate capacitor stores charge, and thus, stores energy in the form of electric potential energy. The total energy sto

DIA [1.3K]

Answer:

The charges on the plates,

Explanation:

A capacitor is an electronic device that can be used for storing of charges. A parallel plate capacitor consists of two plates of equal area separated by a dielectric constant. The energy stored in the capacitor is in the form of potential energy which comes into play during the discharging process of the capacitor.

The energy stored depends majorly on the voltage, area of the plates, distance between the plates and the nature of the dielectric constant of the material between the plates. But it does not depend on the charges on the plates.

7 0

3 years ago