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3 years ago

NEED AN ANSWER FROM SOME ONE THAT IS FROM K12

Physics

2 answers:

marshall27 [118]3 years ago

4 0

Answer:

1. Planetary Model

2. Protons and Neutrons

Explanation:

1. planetary model is also known as quantum model, which is the model we use nowadays

2. Majority of the mass is from the total of neutrons and protons, electrons are so small that it is not included with the mass

Hope this helps

TiliK225 [7]3 years ago

4 0

Answer: 1st one is: Planetary Model

2nd one is: Protons and Neutrons

Explanation:

I took this quiz today, i hope this helps have a great day! :)

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How can i solve this?

Luba_88 [7]

El peso anywhere = (mass) x (gravity there). Gravity en la Tierra = 9.81 m/s^2. Gravity en la Luna = 1.62 m/s^2.

4 0

3 years ago

A cart starts at x = +6.0 m and travels towards the origin with a constant speed of 2.0 m/s. What is it the exact cart position

Ira Lisetskai [31]

Answer:

At the origin (x' = 0 m)

Explanation:

Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.

The formula of speed is given as

S = d/t................. Equation 1

Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.

make d the subject of the equation,

d = St ................. Equation 2

Given: S = 2.0 m/s, t = 3.0 s

Substitute into equation 2

d = 2(3)

d = 6 m.

From the above, the cart covered a distance of 6 m in 3 s.

The exact position of the cart = Initial position-distance covered

x' = x-d............ Equation 3

Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.

Given: x = +6.0 m, d = 6 m.

Substitute into equation 3

x' = +6-6

x' = 0 m.

Hence the cart will be at 0 m (origin) 3 s later

5 0

3 years ago

if a pressure of 70.kPa on a volume of 80.cm cubed is reduced to 10.kPa, by what factor does the pressure change

boyakko [2]

70-10/70 x 100 percentage change ....

60/70, 6/7 fract change

6 0

3 years ago

I did questions 2 and 3 but I don't know if they are right. Someone help!

Tems11 [23]

You got it right my friend

8 0

3 years ago

Read 2 more answers

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what unc

Tpy6a [65]

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ $_{y}$ Δ $_{p}$ $= \frac{h}{2\pi }$

The momentum P $_{y}$ = (mass of electron)(speed of electron)

= $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

= $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

= $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

Δ $_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

Δ $_{y} \geq 1.163 * 10^{-5}m$

5 0

4 years ago