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goldfiish [28.3K]
1 year ago
15

For general projectile motion, the horizontal component of a projectile's velocity aSelect one:O a. continuously increases.O b.

continuously decreases.O c. is zeroO d. remains a non-zero constant.O e. any of the above, depending on position.
Physics
1 answer:
liubo4ka [24]1 year ago
3 0

ANSWER:

d. remains a non-zero constant.

STEP-BY-STEP EXPLANATION:

If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant

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The atmosphere is held together by
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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
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Explanation:

Given that,

Diameter = 10 cm

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Speed v_{1}= 2\ m/s

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(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

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(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})

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(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

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Answer:

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<h2>Please mark me as brainliest by clicking on the crown at the bottom of my answer</h2>

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