1. In a chemical reaction, the forward rate is greater than the reverse rate. Which statement about the reaction is correct?

What kinds of space and matter can light travel through

dem82 [27]

Answer:

Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. (Sound, on the other hand, must travel through a solid, a liquid, or a gas.)

Explanation:

3 0

3 years ago

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The theory of light has been expounded differently by scientists in different periods of time. What is an important reason for t

wlad13 [49]

Technological advances clearly.

Theories change due to other peoples view on the matter and how they experiment with it, but Technology advancements are the greatest factor of all. I mean seriously, we wouldn't know of cells before the microscope and when they first invented it, it advanced further and now we can see things on a whole deeper lever.

8 0

3 years ago

a box is at rest on a ground. An unbalance force acts on the box causes it to start. The box moves 10 meters in 2 seconds and th

nordsb [41]

The box stopped moving because there was a negative acceleration in the x direction caused by friction.
To find the average speed, simply divide the displacement by time: 10 / 2 = 5 m/s

5 0

2 years ago

What is the equivalent resistance of the circuit?

mezya [45]

Answer:

I=VRS=9V90Ω=0.1A

Explanation:

The equivalent resistance is the algebraic sum of the resistances (Equation 10.3. 2): RS=R1+R2+R3+R4+R5=20Ω+20Ω+20Ω+20Ω+10Ω=90Ω. The current through the circuit is the same for each resistor in a series circuit and is equal to the applied voltage divided by the equivalent resistance

5 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.

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3 years ago

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