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3 years ago

The metric unit of force is the

1 answer:

6 0

Newtons. Force is mass times acceleration. Mass is measured in kilograms (kg) and acceleration is measured in meters per second squared (m/s^2.) These units (kg and m/s^2) multiplied together like in the equation force equals mass times acceleration (F=ma) gives a product with Newtons as the unit.

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Planets that are larger in diameter, farther from the sun, and less dense than smaller planets in the solar system are known as

Sergio [31]

Gas giants lol I'm love this kinda stuff nothing else just this question

6 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago

Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc

zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

$Zn + CuSO_4 --> ZnSO_4 + Cu$

so here we have

$Zn = 10.1 g$

$CuSO_4 = 18.6 g$

$ZnSO_4 = 20 g$

$Cu = 8.7 g$

Now total mass of reactant is given as

$M_1 = 10.1 + 18.6 = 28.7 g$

Mass of the product is given as

$M_2 = 20 + 8.7 = 28.7 g$

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

7 0

3 years ago

A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters

julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0

3 years ago

A project for a science class asks teams of students to build a windmill. The windmill is placed in front of a large floor fan.

shtirl [24]

The answer to your question would be team C, because the lifted the most weight in the shortest time. Team A might have been the fastest team but the also lifted the least amount of weight. And team B lifted a good amount of weight but they also did it the slowest.

8 0

3 years ago