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3 years ago

What type of particle found inside the atom is uncharged

Physics

2 answers:

soldi70 [24.7K]3 years ago

3 0

The neutron is uncharged. That's why it was
given a name built on "neu" from "neutral".

tangare [24]3 years ago

3 0

It is an neutron. That is what the answer is.

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The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

6 0

4 years ago

Think about the types of forces holding the atoms together in different chemicals. Are there any patterns that can be drawn from

Leya [2.2K]

Hello. You did not enter the data to which this question refers, which makes it impossible for it to have an exact answer. However, I will try to help you in the best possible way.

The forces that hold the elements together are called intermolecular forces. They are formed by covalent bonds between the molecules and can be called: dipole-induced (occurs between nonpolar molecules that have a negative pole and a positive pole) and dipole-dipole (occurs between polar moileculas, except when hydrogen is present).

5 0

3 years ago

A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp

sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So period of oscillation $T=2\times 3.14\times \sqrt{\frac{3.42}{12}}$

$T=2\times 3.14\times 0.533=3.347sec$

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

3 0

3 years ago

To reach its intended destination on time a cruise ship needs to travel north at 70 mph. However, it is travelling in the Gulf o

bogdanovich [222]

Answer:

speed is 81.03 mph

direction is N 3.58 W

Explanation:

given data

travel north = 70 mph

Stream current = 12 mph

direction = S 25° E

result due north = 70 mph

to find out

speed and direction

solution

we will get component of resultant that is

v cosθ and v sinθ

( 12cos295 , 12 sin295 ) at ( 0, 70)

as that we can say

v sinθ + 12sin295 = 70 ....................1

v cosθ + 12 cos295 = 0 ......................2

vcosθ = -5.0714

vsinθ = 80.8756

now by ratio

cosθ /sinθ = -5.0714/ 80.8756

cot θ = -0.0627

θ = 93.58

so direction is N 3.58 W

and

we know

vcosθ = - 12cos295

v = - 12cos295 / cos(93.58)

v = 81.03 mph

so speed is 81.03 mph

4 0

3 years ago

A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slide

-BARSIC- [3]

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

8 0

4 years ago