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3 years ago

What are the differences between the plum pudding model, nuclear model and the atomic model

2 answers:

8 0

The plum pudding diagram models the protons and electrons in an atom and gives evidence of different charges in an atom. The nuclear model was made to prove that the subatomic particles were placed in different places in the atom such as the neutrons and protons being in the nucleus and the electrons on the outer shells. The atomic model, buy Dalton, was made to prove that the atom is always in constant motion and never still in one place.

HOPE THIS HELPED :)

kakasveta [241]3 years ago

7 0

The plum pudding model suggested that the electrons were dispersed throughout the atom (like chocolate chips in a cookie) and the space was positively charged so that in the end the atom was neutral. Today, people know that the electrons are in a "cloud" around the atom and the protons (and neutrons) are in the nucleus, at the center of the atom.

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How many significant figures are there in : (a) 0.000054 (b) 3.001 x 10^5 (c) 5.600

melomori [17]

Answer:

(a) 2 (b) 4 (c) 4

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

0.000054 has 2 significant figures.

3.001 x 10⁵ has 4 significant figures.

5.600 has 4 significant figures.

4 0

3 years ago

Is this right? Please tell me why its wrong or right

madam [21]

B,A,D,C u can check this by using formula of momentum P=mv..

8 0

2 years ago

A stone tied at one end of the string is whirled in vertical circle, where is the tension in the string be maximum and minimum?

Ber [7]

<h3>answer</h3>

The minimum velocity required for a object to rotate in a verticle plane is v

gr.

So, at bottom point, T

max

=mv

/r+mg=6mg.

At top-most point velocity is v

=gr.

So, T

min

=mv

/r−mg=0

So T

max

−T

min

=6mg=2×g⇒m=1/3 kg

4 0

3 years ago

What is a travelling wave and a standing wave? What are the differences between both of them?

solniwko [45]

What is a travelling wave and a standing wave? What are the differences between both of them?

Answer: First of all we have to understand that a traveling wave is an organized disturbance traveling with a well defined wave speed. On the other hand standing waves are the combination of period waves with their reflected waves creating double sided waves. The differences between them is that standing waves have nodes and antinodes while a traveling wave does not.

I hope it helps, Regards.

7 0

3 years ago

Read 2 more answers

If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago