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3 years ago

An electron has _____

Physics

2 answers:

Soloha48 [4]3 years ago

8 0

Answer:

option (D)

Explanation:

The elementary particles of atom are protons, neutrons and electrons. The protons and neutrons are residing in the nucleus and electron revolves around the nucleus in fixed stationary orbits.

The mass of proton and a neutron is almost same and equal to 1.67 x 10^-27 Kg.

The mass of an electron is 9.1 x 10^-31 Kg.

The mass of electron is much less than the mass of proton or a neutron.

Proton is 1837 times heavier than the electron.

Marrrta [24]3 years ago

3 0

a proton and neutron weigh about the same, while an electron is about 1/2000 of the weight of either a proton or neutron

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When you pick a book up off the floor you are tranfering __________ energy from your hands to the ____________ energy of the boo

MakcuM [25]

When you pick a book off the floor you are transferring kinetic energy from your hands to the potential energy of the book.

3 0

4 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

3 years ago

How much energy, in joules, is released when 17 kilograms of granite is cooled from 45°C to 21°C?

zepelin [54]

The final answer is -322, 320 Joules. The solution for the problem is:

The equation that must be used in this problem is:

U = mCp(Tf-Ti)

where:
U = energy released or absorbed in Joules
m = mass in kg
Cp = specific heat of material in J/kg-C
Tf = final temperature, C
Ti = initial temperature, C

Looking up the Cp of granite gives a value of 790 J/kg-C.

U = 17kg (790 J/kg-C) (21-45)
<span>U = - 322, 320 Joules (negative means heat is released)</span>

4 0

4 years ago

Read 2 more answers

A mass of 2.1 kg the ball leaves her hand with the speed of 30 mass find the energy of the ball

snow_lady [41]

Ke= 1/2 x m x v^2
Ke= 1/2 x 2.1 x 30^2
Energy = 945 J

8 0

3 years ago

Suppose you wanted to be able to see astronauts on the moon. what is the smallest diameter of the objective lens required to res

joja [24]

r = distance of moon from earth = 3.84 x 10⁸ m

R = size of the object on moon = 0.67 m

D = diameter of the lens = ?

$\lambda$ = wavelength of light = 550 x 10⁻⁹ m

Using the equation for Rayleigh criterion

R/r = 1.22 $\lambda$ /D

inserting the values

0.67/(3.84 x 10⁸) = 1.22 (550 x 10⁻⁹) /D

D = 384.6 m

4 0

3 years ago