HEY GUYS IS THIS TRUE OR FALSE????????

Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra

pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

$F = mg$

now the acceleration is given as

$F = ma = mg$

$a = g$

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

$y = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2y}{g}}$

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

3 0

3 years ago

During energy transformation, all energy systems: are

Tom [10]

B: Energy lose

i say this because in order to change they lose energy.

3 0

3 years ago

Read 2 more answers

Earth's subsystems are the geosphere, atmosphere, hydrosphere, and biosphere. Energy and matter are transferred when Earth's sub

Leona [35]

Answer:

ok

Explanation:

The Earth is made of several subsystems or "spheres" that interact to form a complex and continuously changing whole called the Earth system. Scale

Processes operating in the Earth system take place on spatial scales varying from fractions of millimeters to thousands of kilometers, and on time scales that range from milliseconds to billions of years.

Examples of instantaneous - breathing; rotation of the Earth; earthquake

Examples of long term - making coal; plate tectonics

Cycles

The Earth system is characterized by numerous overlapping cycles in which matter is recycled over and over again. Cycles involve multiple spheres and systems interactions.

Examples of cycles: day and night; rock cycle; seasons

Energy

The Earth system is powered by energy from two major sources: the Sun and the planet's internal heat.

Humans and the Earth System

People are part of the Earth system and they impact and are impacted by its materials and processes.

6 0

2 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

If an object has a mass of 20 grams and a volume of 40 cm3, what is its density in g/cm3?

sattari [20]

Density = Mass/Volume
So, given mass = 20 g and volume = 40 cm^3
By substituting in above equation, Density = 20/40 = 0.5 g/cm^3
Hope it helps.

3 0

3 years ago