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3 years ago

Two models of the same compound are shown. In what way is Model A better than Model B?

Physics

2 answers:

Murljashka [212]3 years ago

8 0

Answer:

C- model a shows the three demensiobal shape of the molecule but model b does not

Explanation:

just did the quiz

astra-53 [7]3 years ago

7 0

The correct answer is Model A shows the three-dimensional shape of the molecule, but Model B does not.

Explanation:

Model A and B show the structure of a molecule. In the case of model A, the structure is represented through the use of three-dimensional shapes, while in model B the structure is represented using the letters of each element and showing how each element is connected to others.

In this context, one feature that makes model A better is that this represents the molecule using a 3D model, which is better to understand how the molecule looks like and what is its structure. Moreover, both models are alike because they show the number of atoms of each element, although model A does not show the types of elements.

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our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three

Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The torque produced by the pile of rocks is $\tau = 35.63\ N \cdot m$

The distance of the single for equilibrium to occur is $r_s =1.62 \ m$

Explanation:

From the question we are told that

The mass of the left rock is $m_s = 2.25 \ kg$

The mass of the rock on the right $m_p = 10.1 kg$

The distance from fulcrum to the center of the pile of rocks is $r_p = 0.360 \ m$

Generally the torque produced by the pile of rock is mathematically represented as

$\tau = m_p * g * r_p$

Substituting values

$\tau = 10.1 * 9.8 * 0.360$

$\tau = 35.63\ N \cdot m$

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

The torque due to the single rock is

$\tau = m_s * g * r_s$

At equilibrium the both torque are equal

$35.63 = m_s * r_s * g$

Making $r_s$ the subject of the formula

$r_s = \frac{35.63 }{m_s * g}$

Substituting values

$r_s = \frac{35.63 }{2.25 * 9.8}$

$r_s =1.62 \ m$

6 0

3 years ago

If you were trying to describe the difference between power and work you could say:

avanturin [10]

Power is the energy transferred or "WORK DONE" in one second

6 0

2 years ago

Read 2 more answers

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

2 years ago

A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

dalvyx [7]

Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

6 0

3 years ago

Suppose there are 100,000 atoms of a radioactive substance that has a ½ life of 10 minutes. How many atoms will remain after 40

dezoksy [38]

Answer:

c. 12,500

Explanation:

Original number of atoms = 100,000 atoms

Half- life = 10min

Unknown:

The number of atoms that will remain after 10min = ?

Solution:

The half - life is the time taken for half of a radioactive substance to decay by half.

Time taken Number of atom half life

10min 100000 _

20min 50000 1

30min 25000 2

40min 12500 3

6 0

2 years ago