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2 years ago

(d) शून्य

2 answers:

8 0

The answer is a 20 or 140 u choose because u subtract or add these numbers and sorry I can’t speak sri lanka or Indian but I know how to read them

Travka [436]2 years ago

8 0

^^^^^^^^^^^^^^%^no se

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A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

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2 years ago

Can viruses, bacteria, and fungi be considered parasites?

Pani-rosa [81]

Yes, they live off of other organisms and harm the organisms.

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2 years ago

You have a stopped pipe of adjustable length close to a taut 62.0-cm, 7.25-g wire under a tension of 3910 N . You want to adjust

Ber [7]

Answer:

L = 6 cm

Explanation:

Second overtone of the wire is same as third harmonic

so its frequency is given as

$f = \frac{3}{2L}\sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{M}{L}$

$\mu = \frac{7.25 \times 10^{-3}}{0.62}$

$\mu = 0.0117 kg/m$

now we have

$f = \frac{3}{2\times 0.62}\sqrt{\frac{3910}{0.0117}}$

$f = 1398.6 Hz$

Now fundamental frequency of sound in a pipe is given as

$f = \frac{v}{4L}$

$1398.6 = \frac{340}{4L}$

$L = 0.06 m$

$L = 6 cm$

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2 years ago

At which point is the object not moving?

Alenkinab [10]

It’s none of those because it’s moving at a constant rate

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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us

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3) Earth is about 150 million km from the Sun, and the apparent brightness of the Sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the Sun if we were located five times Earth's distance from the Sun. Answer: The Sun would appear 1/25 times as bright.

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3 years ago

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