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timama [110]
3 years ago
9

Places on earth where most of the earthquakes originated or some mountains and

Physics
1 answer:
BlackZzzverrR [31]3 years ago
5 0

Answer:

techtonic

Explanation:

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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no
Rzqust [24]

Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = A_{y} +  B_{y} + C_{y}

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0
3 years ago
A force F=0.12N is aplied on spring and spring elongates by 3cm . specific constant of spring ​
PilotLPTM [1.2K]

The spring constant is 4 N/m

Explanation:

When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

F=kx

where

F is the magnitude of the spring applied

k is the spring constant

x is the elongation of the spring, relative to its equilibrium position

For the spring in this problem, we have:

F = 0.12 N (force applied)

x = 3 cm = 0.03 m (elongation of the spring)

Therefore, we can solve the formula for k to find the spring constant:

k=\frac{F}{x}=\frac{0.12}{0.03}=4 N/m

Learn more about forces:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

4 0
3 years ago
Express the kinetic energy K in terms of the potential energy U.<br><br><br> K=GMm/2R
max2010maxim [7]

Answer:

K = -½U

Explanation:

From Newton's law of gravitation, the formula for gravitational potential energy is;

U = -GMm/R

Where,

G is gravitational constant

M and m are the two masses exerting the forces

R is the distance between the two objects

Now, in the question, we are given that kinetic energy is;

K = GMm/2R

Re-rranging, we have;

K = ½(GMm/R)

Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;

K = -½U

7 0
3 years ago
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
Sally Sizzle builds a circuit by connecting three resistors in parallel with a 9V battery. The current in this circuit is 3.6 am
Andru [333]

Then answer is C 9 ohms

7 0
3 years ago
Read 2 more answers
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