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3 years ago

A motorboat and a pwc are meeting head-on. which one is the stand-on vessel?

2 answers:

8 0

Based on the situation given, there could be a significant possibility that neither on the motorboat nor the PWC that could be considered as the stand-on vessel. It is because these water vehicles have the ability to consider using their respective "right-of-way" so as to avoid a collision.

Rina8888 [55]3 years ago

6 0

Explanation:

A PWC is a small motorboat with greater maneuverability. Both the motor boat and the PWC is a power driven boat which can sail with more speed and horse power. A PWC is small in size as compared to the motorboat. So the boat which is smaller and with greater maneuverability must give way to the other boat to avoid collision. So the motorboat must be a stand on vessel.

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2 years ago

Leticia timed how fast five apple slices turned brown (oxidate) after being being dipped in different preservatives such as lemo

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Answer:sorry but i no understand

Explanation:

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3 years ago

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A test tube is in uniform circular motion in a centrifuge, which is a piece of scientific equipment used to separate solids from

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Since direction is part of velocity, and the object is moving in a circle, its velocity is constantly changing.

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2 years ago

A horizontal pipe 18.0 cm in diameter has a smooth reduction to a pipe 9.00 cm in diameter. If the pressure of the water in the

Rzqust [24]

Answer:

The rate of flow of water is 71.28 kg/s

Solution:

As per the question:

Diameter, d = 18.0 cm

Diameter, d' = 9.0 cm

Pressure in larger pipe, P = $9.40\times 10^{4}\ Pa$

Pressure in the smaller pipe, P' = $2.80\times 10^{4}\ Pa$

Now,

To calculate the rate of flow of water:

We know that:

Av = A'v'

where

A = Cross sectional area of larger pipe

A' = Cross sectional area of larger pipe

v = velocity of water in larger pipe

v' = velocity of water in larger pipe

Thus

$\pi \frac{d^{2}}{4}v = \pi \frac{d'^{2}}{4}v'$

$18^{2}v = 9^v'$

v' = 4v

Now,

By using Bernoulli's eqn:

$P + \frac{1}{2}\rho v^{2} + \rho gh = P' + \frac{1}{2}\rho v'^{2} + \rho gh'$

where

h = h'

$\rho = 10^{3}\ kg/m^{3}$

$9.40\times 10^{4} + \frac{1}{2}\rho v^{2} = 2.80\times 10^{4} + \frac{1}{2}\rho (4v)^{2}$

$6.6\times 10^{4} = \frac{1}{2}\rho 15v^{2}$

$v = \sqrt{\frac{2\times 6.6\times 10^{4}}{15\times 10^{3}}} = 8.8\ m/s$

Now, the rate of flow is given by:

$\frac{dm}{dt} = \frac{d}{dt}\rho Al = \rho Av$

$\frac{dm}{dt} = 10^{3}\times \pi (\frac{18}{2}\times 10^{- 2})^{2}\times 8.8 = 71.28\ kg/s$

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2 years ago

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Two charges and are separated by 10cm on the x- axis. Q2 is at the origin and Q1 is 12cm to the right.

Evgen [1.6K]

Answer:

Explanation:

Q₂ is at the origin and Q₁ is at 12 cm on x -axis

Force between Q₁ and Q₂

= K x Q₁ x Q₂ / .12 where k is a constant equal to 9x 10⁹

Force F = 9 x 10⁹ x Q₁ x Q₂ / .12²

= 625 x 10⁹ x Q₁Q₂ N

This is the force each one exerts on other . So electric force on Q₂ due to Q₁ is 625 x 10⁹ x Q₁Q₂ N . Its direction is towards left as both will repel each other.

Electric field due to charge Q₂ at - 2 cm

= 9 x 10⁹ x Q₂ / .02²

= 22500 x 10⁹ x Q₂ N/C , towards left .

Electric field due to charge Q₁ at - 2 cm

= 9 x 10⁹ x Q₁ / .14²

= 459.18 x 10⁹ x Q₁ N/C , towards left

As both the fields are to the same direction

Net field = 75 x 10⁹ x Q₂ + 459.18 x 10⁹ x Q₁

10⁹ ( 22500 x Q₂ + 459.18 x Q₁ ) N/C .

Direction will be towards left.

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2 years ago