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fredd [130]
4 years ago
10

What action involves a chemical change?

Physics
1 answer:
Likurg_2 [28]4 years ago
5 0
<span>BAKING A CAKE </span>

Chemical change is a process where a current substance changes or is made into a new type of substance
. Unlike the physical change, which is reversible. Chemical change stays into a its new form. Take for instance these -physical change- examples, making ice cubes. The process involves solidification or freezing where the water becomes ice or solid but when it melts back to its original or typical form with respect to temperature, it’s still water. When the paper is cut into pieces it isn’t burned or exposed to a stimuli that can trigger immediate change in its composition. It’s still the same. On the contrary, baking a cake involves these different compositions or substances –flour, egg, yeast and etc. that is baked to a cake, a newly formed unified substance of all the included ingredients.      



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A physical quantity X is connected from X = ab2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3
Hoochie [10]

Answer: 11%

Explanation:

Given that

X = ab^2/C. Calculate percentage error in X, when percentage error in a,b,c are 4,2 and 3 respectively.

Percentage error of b = 2%

Percentage error of b^2 = 2 × 2 = 4

When you are calculating for percentage error that involves multiplication and division, you will always add up the percentage error values.

Percentage error of X will be;

Percentage error of a + percentage error of b^2 + percentage error of c

Substitute for all these values

4 + 4 + 3 = 11%

Therefore, percentage error of X is 11%

3 0
3 years ago
Calculate the magnitude of the force between two 3.60 mC point challenges 9.3 cm apart
vladimir1956 [14]

Q = magnitude of charge on each of the two point charge = 3.60 mC = 3.60 x 10⁻³ C

r = distance between the two point charges = 9.3 cm = 0.093 m

k = constant = 9 x 10⁹ Nm²/C²

F = magnitude of the force between the two point charges = ?

according to coulomb's law , force between two charges is given as

F = k Q²/r²

inserting the values

F = (9 x 10⁹) (3.60 x 10⁻³)²/(0.093)²

F = 1.35 x 10⁷ N

3 0
3 years ago
If x2 = 60, what is the value of x? plus or minus square root 30 plus or minus square root 60 ±30 ±120
Zielflug [23.3K]
X2 = 60
/ 2 / 2
x = 30
Plus or minus square root 60
4 0
3 years ago
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

3 0
3 years ago
What is the net work done on the 20kg block while it moves the 4 meters?
Vinil7 [7]

Answer:

The answer to your question is 784.8 J. None of your answer, did you forget some information?

Explanation:

Data

mass = 20 kg

distance = 4 m

work = ?

Formula

Work = force x distance

Force = mass x gravity

Process

1.- Calculate the weight of the block

     Weight = 20 x 9.81

     Weight = 196.2 N

2.- Calculate the work done

     Work = 196.2 x 4

     Work = 784.8 J

5 0
4 years ago
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