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3 years ago

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A bar of gold has a temperature of 31°C, and a bar of aluminum has a

1 answer:

Katarina [22]3 years ago

3 0

Different dense matters

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A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?

ser-zykov [4K]

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

${ \rm{ {v}^{2} = {u}^{2} + 2gs}}$

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

${ \rm{0 {}^{2} = {2}^{2} - (2 \times 9.8 \times s)}} \\ \\ { \rm{4 = 19.6s}} \\ \\ { \rm{s = 0.204 \: m}}$

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2 years ago

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Phosphate never enters the<br> O atmosphere<br> O ground<br> O water<br> O ocean

gizmo_the_mogwai [7]

I think you can google this because I really don’t know the answer I’m so sorry

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3 years ago

How can we show the magnetic force of a magnet illustrate with an example

morpeh [17]

Answer:

take some sharp mixture of iron spread it from up the magnet you will see the magnetic field of magnet

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3 years ago

Your performance on the pacer is used to measure which components of health- related fitness?

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Your performance on the pacer test is used to measure cardiovascular components of health-related fitness

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3 years ago

A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above

Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed $v_{1}= 2\ m/s$

Speed $v_{2}=7\ m/s$

Pressure in main pipe $P_{1}=2\times10^{5}\ Pa$

(I). We need to calculate the diameter

Using equation of continuity

$Av_{1}=Av_{2}$

$\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}$

$(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7$

$d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}$

$d_{2}=5.345\ cm$

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

$P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})$

$P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)$

$P_{2}=1.28500\times10^{5}\ Pa$

(III). If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}$

$P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})$

Here, $h_{3}=0$

Put the value in the equation

$P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17$

$P_{3}=3.5400\times10^{5}\ Pa$

Hence, This is required solution.

7 0

3 years ago