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3 years ago

The minimum charge on any object cannot be less than

Physics

1 answer:

BigorU [14]3 years ago

7 0

Answer:

1.6 x 10^{-19} Coulombs

Explanation:

In Physics, the standard unit of measurement of a charge is Coulombs and it's denoted by C. Also, the symbol for denoting a charge is Q.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

The minimum charge on any object such as an electron cannot be less than 1.6 x 10^{-19} Coulombs and it's usually referred to as the fundamental unit of charge.

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a cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5m/s2 for 8.0

Andreas93 [3]

The final velocity is +15.0 m/s

Explanation:

The motion of the cart is a uniformly accelerated motion (=at constant acceleration), therefore we can use the following suvat equation:

$v=u+at$

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

t is the time

For the cart in this problem, we have:

u = +3.0 m/s (initial velocity)

$a=1.5 m/s^2$ (acceleration)

t = 8.0 s (time)

Substituting, we find the final velocity:

$v=3.0+(1.5)(8.0)=+15.0 m/s$

Learn more about accelerated motion:

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4 0

3 years ago

The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength

VARVARA [1.3K]

So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6

8 0

3 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

3 years ago

Compared to the thickness and density of the continental crust of South America, the oceanic crust of the Pacific floor is______

chubhunter [2.5K]

Answer:

The pacific floor (oceanic crust) is thinner and more denser

Explanation:

The continental crust have a thickness of about 35 to 40 km on an average, and are composed of rocks that has less denser granitic minerals such as feldspar, quartz.

On the other hand, the oceanic crust have a thickness of about 7 to 10 km on an average, and it is comprised of denser mafic rocks that contains high amount of olivine and pyroxene minerals.

Due to this, the oceanic crust subducts below the continental crust during the time of collision.

Thus, the pacific floor (oceanic crust) is thinner and more denser, in comparison to the South American continental crust.

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3 years ago

3/10

lyudmila [28]

Answer:

4,800?

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7 0

2 years ago

The minimum charge on any object cannot be less than​

The minimum charge on any object cannot be less than