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3 years ago

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Suppose you first walk 12.0 m in a direction 200 west of north and then 20.0 m in a direction 40.00 south of west. How far are y

ou from your starting point, and what is the compass direction of a line connecting your starting point to your final position?

1 answer:

Tpy6a [65]3 years ago

7 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

the compass direction of the resultant displacement is $\theta =4.7^o$ south of west

Explanation:

Generally using cosine we can obtain the resultant R as follows

$R^2 = A^2 + B^2 -2ABcos(70)$

=> $R = \sqrt{12^2 + 20^2 - 2(12 ) * (20) cos 70}$

=> $R = 19.48 \ m$

We can obtain the direction of the resultant by first using sine rule to obtain angle C as follows

$\frac{A}{sin C} = \frac{R}{sin70 }$

=> $C= sin ^{-1} [\frac{A * (sin 70)}{R} ]$

=> $C = sin ^{-1} [\frac{20 * (sin 70)}{19.48} ]$

=> $C = 74.7 ^o$

Then the direction is obtained as

$\theta = C - 70$

=> $\theta = 74.7 - 70$

=> $\theta =4.7^o$

Hence the compass direction of the resultant displacement is $\theta =4.7^o$ south of west

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(URGENT) A ball rolls off a ledge. Its velocity is 7.70m/s in a horizontal direction. It falls on the floor 1.60m below the ledg

elena-s [515]

Answer:

the horizontal distance is 4.355 meters

Explanation:

The computation of the horizontal distance while travelling in the air is shown below:

Data provided in the question is as follows

Velocity = u = 7.70 m/s

H = 1.60 m

R = horizontal direction

Based on the above information

As we know that

R = u × time

where,

Time = $\sqrt{\frac{2H}{g} }$

So,

= $7.70\times \sqrt{\frac{2\times 1.60}{10} }$

= 4.355 meters

hence, the horizontal distance is 4.355 meters

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3 years ago

To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,

bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1 = 800 Hz

harmonic frequency f2 = 1120 Hz

harmonic frequency f3 = 1440 Hz

solution

first we get here fundamental frequency that is express as

2F = f2 - f1 ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength = Speed ÷ Fundamental frequency ................2

here speed of waves in air = 340 m/s

so put here value

Wavelength =340 ÷ 160

Wavelength = 2.125 m

so

Length of the pipe will be

Length of the pipe = 0.25 × wavelength ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

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2 years ago

Describe a situation where you add heat to a substance or material but there is no change in temperature. What does this look li

Alisiya [41]

ONE CAN perform this by doing an ideal experiment
by creating an isothermal system
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3 years ago

In any given wave, when the frequency of the wave doubles (f = 2f), which of the following other changes would also take place?

Zanzabum

Answer:

The correct answer is -

B. The velocity would double (v = 2v).

C. The wavelength would be half (λ = λ/2).

Explanation:

A wave has a speed or velocity that is related to the wavelength of the wave and the frequency of the wave and this relationship can be represented by the following equation-

Wave velocity V = Wavelength (λ) * Frequency (f)

Frequency (f) = Velocity (V) / Wavelength(λ).

The frequency and wavelength are inversely proportional and frequency and velocity are directly proportional to each other.

So, if f = 2f then,

putting value in the formula,

2f = 2v/λ, which means, f = 2v and f = λ/2

when the frequency is doubled, the wavelength will be halved and velocity will be doubled.

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2 years ago

You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.

Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

$F_{a} =\frac{F_{f}+F_{i} }{2}$ Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx : displacement

$W = F_{a} (x_{f} -x_{i} )$ :Formula (3)

$x_{f}$ : final deformation

$x_{i}$ :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

$F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N$

$F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N$

We calculate average force applying formula (2):

$F_{a} =\frac{15.4N+6.2N}{2} = 11 N$

We calculate the work done on the spring applying formula (3) : :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

$E_{pf} =\frac{1}{2} *k*x_{f}^{2}$

$E_{pi} =\frac{1}{2} *k*x_{i}^{2}$

$E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J$

$E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J$

W=ΔEp= 5.39 J-0.99 J = 4.4J

:

4 0

2 years ago