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3 years ago

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1. Water flows through a hole in the bottom of a large, open tank with a speed of 8 m/s. Determine the depth of water in the tan

k. Viscous effects are negligible.

1 answer:

lora16 [44]3 years ago

5 0

Answer:

3.26m

Explanation:

See attached file

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A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i

victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

$F= \frac{kqQ}{r^2}$

Where,

$k = 8.99 * 10^9 N m^2 / C^2.$

q = charges of the objects

r= distance/radius

Our values are previously given, so

$q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59$

Replacing,

$F=\frac{kqQ}{r^2}$

$F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}$

$F= 4.8423N$

The force acting on the block are given by,

$F-mgsin\theta = ma$

$a = \frac{F-mgsin\theta}{m}$

$a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}$ $a = 10.31m/s^2$

Therefore the box is accelerated upward.

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3 years ago

Can the reaction distance be a negitive?

leonid [27]

Yes. Displacement is a vector, as opposed to distance, which is a scalar. Since displacement is a vector quantity, the negative sign represents direction. As with velocity and acceleration, it just comes down to how you define your coordinate system

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3 years ago

A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: brainly.com/question/25959744

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The answer is ...
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