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mr Goodwill [35]
2 years ago
4

Charlie is balancing an equation. She has identified the atoms and counted the number of each in the reactants and products. Wha

t would Charlie adjust to make the number of atoms in the reactants the same as the number of atoms in the products?
A: coefficients
B: subscripts
C: superscripts
D: parentheses
Physics
2 answers:
IrinaVladis [17]2 years ago
6 0

Answer;

A: coefficients

Explanation;

The Law of Conservation of Mass states that matter cannot be created or destroyed. Thus, the mass of substances produced in a chemical reaction is always equal to the mass of reacting substances. Therefore, there is need to have the same number of each type of element on each side of a chemical equation.

Coefficients of products and reactants are used to balance the number of atoms of an element on both sides of a chemical equation.


OLga [1]2 years ago
5 0
The answer to your question is A. Coefficients. 

*Hope this helps you! Feel free to message me if you have any further questions. Have a nice day! :)*
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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
2 years ago
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a tur
GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
2 years ago
Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o
sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

                                   T=\frac{2\pi }{\omega}

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
2 years ago
A cylinder is sliced in half along its diagonal. Determine the location of the center of mass and the inertia properties relativ
Black_prince [1.1K]

Answer:

See the explanation below

Explanation:

To better understand this problem, a cylinder sketch is attached before and after the cut, we see that after the cut, the shape of this resembles that of a right triangle.

We can find, the centroid in the xy plane, knowing that the centroid for a triangle is located a third of its base.

In the z axis there is no displacement of the centroid.

3 0
2 years ago
You are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform m
mamaluj [8]
The correct answer is:
<span>paramagnetism 

In fact, paramagnetic materials, when they are placed in a magnetic field, they form an internal magnetic field parallel to the external one and in the same direction. However, unlike ferromagnetic materials, they do not retain their magnetization, so when the external magnetic field is removed, their internal induced magnetic field disappears.</span>
7 0
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