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3 years ago

A hockey puck wont slide very far on concrete or asphalt, but it will slide for a very long time on ice. Why is this?

Physics

2 answers:

Juliette [100K]3 years ago

8 0

Answer:

Concrete and Asphalt are much rougher surfaces and have a higher coefficient of friction because they generate more resistance to an object in motion. Ice has less friction and less energy is lost to heat, causing the puck to slide much further.

Ilya [14]3 years ago

7 0

When hockey players push the puck along the ice it slides causing heat which melts the ice causing the friction against the ice to be less.

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A 256 Hz tuning fork is resonating in a closed tube on a warm day when the speed of sound is 346 m/s. What is the length of the

Ilya [14]

Assuming our "closed tube" is closed at only one end, then
 v = fλ = f*4L/n
 where "n" is the harmonic number. So
 L = nv / 4f = n*346m/s / 4*256Hz = n*0.38 m
 Since the only option in your list that is an integer multiple of 0.38 m is 1.35 m
 I'd say that we're hearing the fourth harmonic.
answer is
A. 1.35 m

5 0

3 years ago

What is meant by the term fitness

Artemon [7]

Answer:

An organisms ability to survive and reproduce in a particular environment.

4 0

3 years ago

Recall that if a line is parallel to the vector v and passes through the point P0, which is the tip of the position vector r0, t

Elenna [48]

Answer:

The vector equation of the line is

$\overrightarrow{r}=+t$

Parametric equations for given line are

$x=7+t\\y=-8+6t\\z=3-13t$

Explanation:

The vector equation of the line is given by

$r(t) = r_{o} + tv$

r₀ = (7, -8, 3)

v = (1, 6, -13)

At these points the vector equation for this line is:

$\overrightarrow{r}=\overrightarrow{r_{o}}+t\overrightarrow{v}\\\overrightarrow{r}=+t$

Parametric equations for given line are

$x=7+t\\y=-8+6t\\z=3-13t$

8 0

3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

7 0

3 years ago

Read 2 more answers

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago