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3 years ago

What is an ampere? NO GUESSING! ANSWERS ONLY!

Physics

2 answers:

Igoryamba3 years ago

7 0

Answer:

the SI base unit of electrical current

Zielflug [23.3K]3 years ago

3 0

Answer:

An ampere is the current that flows through a conductor whose resistance is 1 ohm and the potential difference applied across its ends in 1 volt.

Explanation:

${ \tt{ampere = \frac{1 \: volt}{1 \: ohm} }}$

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With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

hichkok12 [17]

(2.0 x 10² N/Coul) x (-1.6 x 10⁻¹⁹ Coul) = -3.2 x 10⁻¹⁷ N

The magnitude is 3.2 x 10⁻¹⁷ Newton.

3 0

3 years ago

What does it mean, if the specific gravity of an object is less than one?

JulijaS [17]

Answer:

Specific gravity can be used to determine if an object will sink or float on water. ... If an object or liquid has a specific gravity greater than one, it will sink. If the specific gravity of an object or a liquid is less than one, it will float.

hope this helps, have a great day/night, and stay safe!

3 0

2 years ago

Read 2 more answers

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the

Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I  is 1360 W/m²

P sun = 2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars = 3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

6 0

3 years ago

What are the two ways to create plasma?

dimaraw [331]

Answer:

Plasma can be artificially generated by heating a neutral gas or subjecting it to a strong electromagnetic field to the point where an ionized gaseous substance becomes increasingly electrically conductive.

5 0

3 years ago

Read 2 more answers

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius R, at what radial distances in

eimsori [14]

Answer:

Electric field at radius r inside the solid sphere is

Electric field at radius r between inner radius and outer radius of the shell is

E=0 N/C

Explanation:Given

The radius of the solid sphere is

The charge on the solid sphere is

The inner radius of the shell is

The outer radius of the shell is

The total charge on the shell is

PART(A)

The magnitude of electric field at radius r where \\The volumetric charge density of the solid sphere will be

The charge enclosed by the radius r inside the solid sphere is

rho=

According to gauss law

PART(B)

The electric field at radius r where

The shell is conducting so due to induction of charge

The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell

So,

The charge on the inner surface of the shell is

Due to conservation of the charge on the shell

The charge accumulated on the outer surface of the shell is

The charge enclosed by the radius r where

According to gauss law

Read more on Brainly.com - brainly.com/question/13242041#readmore

3 0

3 years ago