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Allushta [10]
1 year ago
15

could you help with question 5. Your solutions to the word problems in volving Newton's Laws should have the following features:

Each solution should begin with one or more free body diagrams with a sign convention.The solution process should follow the GRASP process.Formulae should always be used in a three-step process: formula with letter variables, substitution of values, and then the answer.Answers should be expressed with the correct number of significant digits and with the direction if required.

Physics
1 answer:
Ganezh [65]1 year ago
3 0

ANSWER:

(a) 20 m/s^2

(b) 250 m

STEP-BY-STEP EXPLANATION:

Given:

Mass (m) = 250 g = 0.25 kg

Force (F) = 20 N

Frictional force (Ff) = 15 N

(a)

The net force on the block is:

\begin{gathered} F_{net}=F-F_f \\ F_{net}=20-15 \\ F_{net}=5\text{ N} \end{gathered}

We know that:

\begin{gathered} F=m\cdot a \\ \text{ We replacing} \\ 5=0.25\cdot a \\ a=\frac{5}{0.25} \\ a=20\text{ m/s}^2 \end{gathered}

(b)

We can calculate the partial distance from the following equation:

initial velocity (u) = 0 (start from rest)

\begin{gathered} s=ut+\frac{1}{2}at^2 \\ \text{ we replacing} \\ s=0\cdot5+\frac{1}{2}\cdot20\cdot5^2 \\ s=250\text{ m} \end{gathered}

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A flock of ducks is trying to migrate south for the winter, but they keep being blown off course by a wind blowing from the west
Minchanka [31]

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}

(see the attached graphic)

We have

  • ducks (relative to wind) = 7.0 m/s in some direction <em>θ</em> relative to the positive horizontal direction, or

\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

  • wind (relative to Earth) = 5.0 m/s due East, or

\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)

  • ducks (relative to earth) = some speed <em>v</em> due South, or

\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)

Then by setting components equal, we have

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

We only care about the direction for this question, which we get from the first equation:

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}

\cos\theta=-\dfrac57

\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

8 0
3 years ago
The sound emitted by bats has a wavelength of 4.13 mm.
blagie [28]

Answer:

the frequency of a wave is equal to the linear speed divided the wavelength. so in equation form.

f = v / l

where f is the frequency

v iss the linear speed

l is the wavelength

f = ( 345 m/s ) / ( 4.38 mm ) ( 1 m / 1000 mm )

f = 78767 per second

Explanation:

hope it helps

8 0
4 years ago
Two strings on a musical instrument are tuned to play at 196 hz (g) and 523 hz (c). (a) what are the first two overtones for eac
Tems11 [23]
(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
f_n = n f_1
where f1 is the fundamental frequency.

So, the first overtone (2nd harmonic) of the string is
f_2 = 2 f_1 = 2 \cdot 196 Hz = 392 Hz
while the second overtone (3rd harmonic) is
f_3 = 3 f_1 = 3 \cdot 196 Hz = 588 Hz

Similarly, for the second string with fundamental frequency f_1 = 523 Hz, the first overtone is
f_2 = 2 f_1 = 2 \cdot 523 Hz = 1046 Hz
and the second overtone is
f_3 = 3 f_1 = 3 \cdot 523 Hz = 1569 Hz

(b) The fundamental frequency of a string is given by
f=  \frac{1}{2L}  \sqrt{ \frac{T}{\mu} }
where L is the string length, T the tension, and \mu = m/L is the mass per unit of length. This part  of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them m_{196}, the mass of the string of frequency 196 Hz, and m_{523}, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
\frac{523 Hz}{196 Hz} =  \frac{ \frac{1}{2L}  \sqrt{ \frac{T}{m_{523}/L} } }{\frac{1}{2L}  \sqrt{ \frac{T}{m_{196}/L} }}
and since L and T simplify in the equation, we can find the ratio between the two masses:
\frac{m_{196}}{m_{523}}=( \frac{523 Hz}{196 Hz} )^2 = 7.1

(c) Now the tension T and the mass per unit of length \mu is the same for the strings, while the lengths are different (let's call them L_{196} and L_{523}). Let's write again the ratio between the two fundamental frequencies
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L_{523}} \sqrt{ \frac{T}{\mu} } }{\frac{1}{2L_{196}} \sqrt{ \frac{T}{\mu} }} 
And since T and \mu simplify, we get the ratio between the two lengths:
\frac{L_{196}}{L_{523}}= \frac{523 Hz}{196 Hz}=2.67

(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them T_{196} and T_{523}. Let's write again the ratio of the frequencies:
\frac{523 Hz}{196 Hz}= \frac{ \frac{1}{2L} \sqrt{ \frac{T_{523}}{m/L} } }{\frac{1}{2L} \sqrt{ \frac{T_{196}}{m/L} }}
Now m and L simplify, and we get the ratio between the two tensions:
\frac{T_{196}}{T_{523}}=( \frac{196 Hz}{523 Hz} )^2=0.14
7 0
3 years ago
Bart awoke screaming, shaking, and sweating in his heart was beating rapidly. Homer and Marge, being good parents, had Bart hook
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Answer: D

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Approximately 10kg

Given that gravity on the Moon has approximately 1/6th of the strength of gravity on Earth, a man who weighs 60kg on Earth would weigh approximately 10kg on the Moon.
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