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3 years ago

14

The amount of _______ that occurs when a sound wave encounters a barrier depends on the wave's length.

1 answer:

Masteriza [31]3 years ago

3 0

Diffraction i believe

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Name and explain two sleeping disorders

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Insomnia and night terrors

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3 years ago

Two people are pushing on a desk to the right with a force 15n each what is the net force on the desk

dybincka [34]

30N to the right direction

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3 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

3 years ago

An object moves along the x-axis according to the equation x = 3.00t2 – 2.00t + 3.00,

Ray Of Light [21]

Explanation:

x = 3.00 $t^{2}$ – 2.00t + 3.00,

Distance of object at 2 second,

x (t=2) = 3(4) - 2(2) +3

x (t=2) = 12-4 +3

x (t=2) = 11 m

Distance of object at 3 second,

x (t=3) = 3(9) - 2(3) +3

x (t=2) = 27 - 6 + 3

x (t=2) = 24 m

a) the average speed between t = 2.00 s and t = 3.00 s,

Average speed = $\frac{Total distance}{ Total time}$

Average speed = $\frac{x (t=2) + x (t=3)}{3}$

Average speed = $\frac{24+11}{3}$

Average speed = $\frac{35}{3}$

Average speed = 11.66 $\frac{m}{s}$

b) the instantaneous speed at t = 2.00 s and t = 3.00 s,

Instantaneous speed = $\frac{dx}{dt}$

Instantaneous speed(v) = 6t - 2 $\left \{ {{t=2} \atop {t=3}} \right.$

Instantaneous speed,v(t=2 to t=3) = 18-2-12+2

Instantaneous speed, v = 6 $\frac{m}{s}$

c) the average acceleration between t = 2.00 s and t = 3.00 s

average acceleration = $\frac{average velocity}{time}$

average acceleration = $\frac{11.66}{3-2}$

average acceleration = 11.66 $\frac{m}{s^{2} }$

d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s

instantaneous acceleration = $\frac{dv}{dt}$

instantaneous acceleration =6

instantaneous acceleration = 6 $\frac{m}{s^{2} }$

e) for x =0

0 = 3.00 $t^{2}$ – 2.00t + 3.00

a = 3, b=-2, c=3

t= $\frac{-b \pm \sqrt{b^{2} - 4ac} }{2a}$

t= $\frac{2 \pm \sqrt{4 - 36} }{6}$

t= $\frac{2 \pm \sqrt{-32} }{6}$

general solution of this equation gives imaginary value. Hence, the given object is not at rest.

7 0

3 years ago

Read 2 more answers

HELP PLZ!!<br> can the kinetic energy of an object be negative? explain

BARSIC [14]

Answer:

No it can not

Explanation: Kinetic energy is the energy of motion so it can not be negative the kinetic energy can only be at a point of "0" which is when its not moving. (I hope this helped) :))

7 0

2 years ago