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tatyana61 [14]
2 years ago
6

A person walks 9 meters to the right and 8 meters to the left what's the distance and displacement​

Physics
2 answers:
nordsb [41]2 years ago
5 0
The distance is 17 and the displacement is 1
saw5 [17]2 years ago
4 0

Answer:

walked: 17 meters

displacement: 1

Explanation:

so you add 9+8 and the displacement is 1 b/c when you moved back those 8 meters you were still 1 meter away from where you where originally standing

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It will go about 45-50 mph.

Explanation:

Depending on what you are hauling.

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Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t
ruslelena [56]

Answer:

x_1= 0.0425m

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

m*g = k*x

With two springs, let, T1 be the tension in each spring,  x1 be the extension of each spring.  The spring constant of each spring is 2k so:

T_1 = 2k*x_1

2T_1 = m*g=4k x_1

Solve to x1

x_1=\frac{m*g}{4k}

x_1=\frac{k*x}{4*k}

x_1=\frac{x}{4}

x_1 = 0.170 / 4

x_1= 0.0425m

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What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?
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= 80 N/40 kg
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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

a.

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

b.

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
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