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3 years ago

A mechanic oils his tools between uses. Explain how this relates to friction andefficiency.

Physics

1 answer:

Mazyrski [523]3 years ago

3 0

Answer:

it reduces friction

Explanation:

if you put oil or any liquid between the tool it will shift and change the shape as much as it needs to. It will smooth the bumps between the tools gears as they squeeze together making the friction reduce. It will slide and that reduces the friction.

i hope this helps im lowkey just guessing tho so sorry if it doesnt. lmk tho

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A car traveling in a straight line has a velocity of 5.0 m/s. After an acceleration of 0.75 m/s/s, the cars velocity is 8.0. In

Bogdan [553]

Vs - velocity on beginning
ve - velocity on ending. You've got:
$v_s = 5 \frac{m}{s} \\ v_e=8 \frac{m}{s} \\ \hbox{Then:} \\ \Delta v=v_e - v_s = 8 \frac{m}{s} - 5\frac{m}{s}=3 \frac{m}{s} \\ a=0,75 \frac{m}{s^2} \\ \hbox{And from formula:} \\ a=\frac{\Delta v}{\Delta t} \qquad \Rightarrow \qquad \Delta t= \frac{\Delta v}{a} \\ \hbox{Substitute:} \\ \Delta t=\frac{3\frac{m}{s}}{0,75 \frac{m}{s^2}}= \frac{3}{\frac{3}{4}} s= 3 \cdot \frac{4}{3} s= 4 s$
So he needed 4 second.

3 0

3 years ago

A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago

Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out

PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is

$\lambda _2 = (6.48,6.52) \ cm$

Explanation:

From the question we are told that

The wavelength of the note is $\lambda = 6.50 \ cm = 0.065 \ m$

The difference in beat frequency is $\Delta f = 17.0 \ Hz$

Generally the frequency of the note played by the guitar that is in tune is

$f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

$f_1 = \frac{343}{0.0065}$

$f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

$\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

$f_2 =f_1 \pm \Delta f$

substituting values

$f_2 =f_1 + \Delta f$

$f_2 = 5276.9 + 17.0$

$f_2 = 5293.9 \ Hz$

The wavelength for this frequency is

$\lambda_2 = \frac{343 }{5293.9}$

$\lambda_2 = 0.0648 \ m$

$\lambda_2 = 6.48 \ cm$

For the second value of the second frequency

$f_2 = f_1 - \Delta f$

$f_2 = 5276.9 -17$

$f_2 = 5259.9 Hz$

The wavelength for this frequency is

$\lambda _2 = \frac{343}{5259.9}$

$\lambda _2 = 0.0652 \ m$

$\lambda _2 = 6.52 \ cm$

8 0

3 years ago

Write a small paragraph on how to determine the speed of sound using the time of flight method.

Makovka662 [10]

Answer:

time of flight of a pulse, and these most often

involve triggering of the measuring oscilloscope

with the signal that generates the sound pulse and

timing the time delay of the pulse picked up by a

conveniently placed microphone45

. Loren Winters

has reported a method similar in principle to the

present one, but which uses a completely different

detection system6

Explanation:

6 0

3 years ago

Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5,

hammer [34]

Answer:

1) 5.52 cm , C) 5.5 cm

Explanation:

When a measurement is carried out, in addition to the value of the magnitude, the error or uncertainty of the measurement must occur, in a direct measurement with an instrument the uncertainty is equal to the appreciation of the instrument.

Uniform see the errors by the number of significant figures days, in this cases they are two decimals for which the appreciation of the instrument ± - 0.01

now we can analyze the measurements made

1) 5.52 cm. Validate. It is a valid measurement is within the uncertainty range

2) 6.63 cm. It does not validate. It is out of the error range

3) 5.5 cm Valid. It is within the given error range,

4) 5.93 cm Not Valid. It is out of the error range.

7 0

3 years ago