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3 years ago

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A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the s

outh bank pointed in a direction 26o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground

1 answer:

tamaranim1 [39]3 years ago

7 0

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

$vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j$

(a) The velocity of boat with respect to ground is

$\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s$

(b) The direction is given by

$tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o$

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4 years ago

Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t

Tamiku [17]

Answer:

Approximately $9.62$ .

Explanation:

$y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0]$ .

$y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)]$ .

Notice that sine waves $y_1$ and $y_2$ share the same frequency and wavelength. The only distinction between these two waves is the $(-0.250)$ in $y_2\!$ .

Therefore, the sum $(y_1 + y_2)$ would still be a sine wave. The amplitude of $(y_1 + y_2)\!$ could be found without using calculus.

Consider the sum-of-angle identity for sine:

$\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)$ .

Compare the expression $\sin(a + b)$ to $y_2$ . Let $a = (4.35\, x - 1270)$ and $b = (-0.250)$ . Apply the sum-of-angle identity of sine to rewrite $y_2\!$ .

$\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Therefore, the sum $(y_1 + y_2)$ would become:

$\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Consider: would it be possible to find $m$ and $c$ that satisfy the following hypothetical equation?

$\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}$ .

Simplify this hypothetical equation:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}$ .

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

$\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}$ .

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real $x$ and $t$ , the following should be satisfied:

$\displaystyle 1 + \cos(-0.250) = m\, \cos(c)$ , and

$\displaystyle \sin(-0.250) = m\, \sin(c)$ .

Consider the Pythagorean identity. For any real number $a$ :

${\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2$ .

Make use of the Pythagorean identity to solve this system of equations for $m$ . Square both sides of both equations:

$\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2$ .

$\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2$ .

Take the sum of these two equations.

Left-hand side:

$\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}$ .

Right-hand side:

$\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}$ .

Therefore:

$m^2 = 2 + 2\, \cos(-0.250)$ .

$m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98$ .

Substitute $m = \sqrt{2 + 2\, \cos(-0.250)}$ back to the system to find $c$ . However, notice that the exact value of $c\!$ isn't required for finding the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ .

(Side note: one possible value of $c$ is $\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125$ radians.)

As long as $\! c$ is a real number, the amplitude of $(y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)$ would be equal to the absolute value of $(4.85\, m)$ .

Therefore, the amplitude of $(y_1 + y_2)$ would be:

$\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}$ .

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3 years ago

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mrs_skeptik [129]

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Read 2 more answers

Nana76 [90]

The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that action and reaction are equal and opposite.

Fa = -Fb

The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.

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Learn more about force here: brainly.com/question/12970081

#SPJ1

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