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3 years ago

• ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.

Physics

1 answer:

nignag [31]3 years ago

5 0

Answer:

45.6 m at $80.5^{\circ}$ south of west

Explanation:

Let's take the north-south direction as y-direction (with south being positive) and east-west direction as x-direction (with west being positive). Therefore, the two components of Cody's motion are:

- $d_y = 45.0 m$ (south)

- $d_x = 7.50 m$ (west)

Since they are perpendicular, the magnitude of the net displacement can be calculated by using Pythagorean's theorem:

$d=\sqrt{d_x^2+d_y^2}=\sqrt{7.50^2+45.0^2}=45.6 m$

The direction instead can be measured as follows:

$\theta = tan^{-1} (\frac{d_y}{d_x})=tan^{-1}(\frac{45.0}{7.50})=80.5^{\circ}$

And given the convention we have used, this angle is measured as south of west.

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Which is a disadvantage of using wind as an energy source?

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Answer:

it produces

Explanation:

a lot of waste

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3 years ago

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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

marshall27 [118]

Answer:

The resultant velocity is $v_t=10 knots$

Explanation:

Apply the law of conservation of momentum

$M_L *v_L + M_f * V_f = (M_L + M_f) v_t$

Where $M_L$ is the mass of the Luxury Liner = 40,000 ton

$v_L$ is the velocity of Luxury Liner = 20 knots due west

$M_f$ mass of freighter = 60,000

$v_f$ is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

$v_f = 0 \ knots$

So the equation would be

$M_L *v_L = (M_L + M_f) v_t$

Substituting values

$40000*20 = (40000+ 60000)v_t_w$

Where $v_t_w$ the final velocity due west

Making $v_t_w$ the subject

$v_t_w = \frac{40,000* 20}{(40000 + 60000)}$

$= 8 \ knots$

Apply the law of conservation of momentum toward the the north direction

$v_L = 0 \ knots$

So the equation would be

$M_f *v_f = (M_L + M_f) v_t_n$

Where $v_t_n$ the final velocity due north

Making $v_t_n$ the subject

$v_t_n = \frac{60,000* 10}{(40000 + 60000)}$

$= 6 \ knots$

The resultant velocity is

$v_t = \sqrt{v_t_w^2 + v_t_n^2}$

$= \sqrt{8^2 +6^2}$

$v_t=10 knots$

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3 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$