Pleasee help me pwease

Describe the mirror formula for a concave mirror in a shortcut way

sattari [20]

Answer:

1/f =1/v+1/u

Explanation:

4 0

3 years ago

PLEASE HELP ASAP Which one of the following is a step used for balancing chemical equations?

scZoUnD [109]

Answer:

C. Count the atoms in each substance in the reactants and products.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

In order for a chemical equation to be balanced, the condition which must be met is that the number of atoms in the reactants equals the number of atoms in the products.

This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

One of the step used for balancing chemical equations is to count the atoms in each substance in the reactants and products.

For example;

NH3 + O2 -----> NO + H2O

The number of atoms in each chemical element are;

For the reactant side:

Nitrogen, N = 1

Hydrogen, H = 3

Oxygen, O = 2

For the product side;

Nitrogen, N = 1

Hydrogen, H = 2

Oxygen, O = 2

When we balance the chemical equation, we would have;

NH3 + 3O2 -----> 4NO + 2H2O

3 0

3 years ago

Drag and drop the terms at the left to match the appropriate descriptions at the right. ResetHelp Visual acuity Emmetropia Accom

Lubov Fominskaja [6]

Answer: Visual acuity: sharpness of vision.

Myopia: nearsightedness

Refraction: bending of light rays.

Emmetropia: normal vision.

Accommodation: changes the shape of the eye lens to focus light on the retina.

Presbyopia: age-related farsightedness due to loss of elasticity in the lens.

Astigmatism: reduction in visual acuity due to changes in the cornea or lens.

Hyperopia: farsightedness

Myopia and hyperopia are refractive errors of the eye.

Presbyopia occurs in old age people.

Explanation:

Visual acuity: It can be defined as the inability to observe the details of shape of the object. Person loses sharpness in vision.

Myopia: It is a defect in vision in which person is able to observe the near by objects clearly but not able to see the distant objects.

Refraction: It can be defined as the bending of beam of light when it passes through from one substance to another.

Emmetropia: It is a vision without any defect.

Accommodation: It is the ability of the eye to adjust its focal length and adjusting the light on focus.

Presbyopia: It can be defined as the loss of ability of eye to focus on the object. It occurs in old age.

Astigmatism: It is a refractive error in which the eye does not focus light on retina.

Hyperopia: It is also called as farsightedness. Distant objects can be seen clearly but nearby appears blurry.

5 0

3 years ago

BMW has introduced a program of covered scheduled maintenance for its cars. That is, when you bring in your vehicle for maintena

Kipish [7]

Answer:

Usage

Explanation:

A product life cycle refers to the duration from the introduction of the product into the market until it's taken off the shelves.

A program of covered scheduled maintenance for the cars refers to the practice in which the dealership will cover the costs of everything when you bring in your vehicle for maintenance.

In this question, the mentioned practice specifically targets the stage - Usage of the product life cycle

6 0

4 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

4 years ago

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