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denis-greek [22]

4 years ago

14

the dog has a momentum of 60 kilogram meters per second west. the dog has a velocity of 3 meters per second west. what is the ma

ss of the dog?

2 answers:

Anestetic [448]4 years ago

7 0

We know that
P=mv so
So dog has 20 kg mass...according to formula

lianna [129]4 years ago

5 0

P=mv=>m=p/v=60/3=20kg

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An automatic shotgun has a mass of 10 kg. It fires 30 g bullets at the rate of 6 bullets per second with a speed of 400 m/s. Wha

igomit [66]

Answer:72 N

Explanation: thats whats right

6 0

3 years ago

A 1.0-m-tall vertical tube is filled with 20°C water. A tuning fork vibrating at 580 Hz is held just over the top of the tube as

Mademuasel [1]

Answer:

water heights of the tube are 0.851 m , 0.553 m, 0.255 m

Explanation:

given data

frequency = 580 Hz

temperature = 20°C

tube = 1 m

to find out

water heights of the tube

solution

we will apply here formula for length that is

length L = v ( 2n -1 ) / 4f

here v is velocity o sound that is 343.2 m/s

so for n = 1

L = 343.2 ( 2(1) -1 ) / 4(580) = 0.147931 m

for n = 2

L = 343.2 ( 2(2) -1 ) / 4(580) = 0.443793 m

for n = 3

L = 343.2 ( 2(3) -1 ) / 4(580) = 0.739655 m

for n = 4

L = 343.2 ( 2(4) -1 ) / 4(580) = 1.035517 m is greater than 1

and so here height is measured less than 1 m

so water heights of the tube are 1 m - 0.147931 m , 1 m - 0.443793 m, 1 m - 0.739655 m

so water heights of the tube are 0.851 m , 0.553 m, 0.255 m

3 0

3 years ago

Who first thought that the sun orbited the earth?

TEA [102]

Answer:

It's Ptolemy

Explanation:

He thought the Earth was the universe's center and everything revolved around us.

4 0

3 years ago

An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise

Viefleur [7K]

Answer:

The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

$\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}$

$\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}$

$f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)$

$\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)$

$\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

$v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

Substitute the given values in the formula,

$v_{s}=343+\frac{262}{271}(343-0)$

$v_{s}=343+0.966(343)$

$v_{s}=343-331.33$

$v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Therefore, The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

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3 years ago

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Kaylis [27]

<span>A particle released during the fission of uranium-235 is a "Neutron"</span>

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3 years ago