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2 years ago

Plss s ss s s s s s s s s

Physics

2 answers:

Crazy boy [7]2 years ago

8 0

Answer:

i font know ewan madaling sabihin yan

hichkok12 [17]2 years ago

7 0

Hey, I’ll may be wrong but here it is!

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What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m

pychu [463]

Complete question:

A taut rope has a mass of 0.123 kg and a length of 3.54 m. What average power must be supplied to the rope to generate sinusoidal waves that have amplitude 0.200 m and wavelength 0.600 m if the waves are to travel at 28.0 m/s ?

Answer:

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

Explanation:

Velocity = Frequency X wavelength

V = Fλ ⇒ F = V/λ

F = 28/0.6 = 46.67 Hz

Angular frequency (ω) = 2πF = 2π (46.67) = 93.34π rad/s

Average power supplied to the rope will be calculated as follows

$P_{avg} =\frac{1}{2} \mu \omega^2 A^2 V$

where;

ω is the angular frequency

A is the amplitude

V is the velocity

μ is mass per unit length = 0.123/3.54 = 0.0348 kg/m

$P_{avg} =\frac{1}{2} ( 0.0348)(93.34 \pi )^2 (0.2)^2 (28)$ = 1676.159 watts

The average power supplied to the rope to generate sinusoidal waves is 1676.159 watts.

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3 years ago

How fast should a girl of 35 kg run so that her kinetic energy becomes 700J<br> of?

Kipish [7]

V^2 = 700J/0.5*35kg

V = square root of 40

v = 6.324 m/s

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2 years ago

Read 2 more answers

Hydrogen is a nonmetal with many nonmetal properties. Why is it on the periodic table with the metals in Group 1?

mrs_skeptik [129]

Answer:

This is because it has ns1 electron configuration like the alkali metals.

Pls Mark Brainiest, am trying to become Virtuoso

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2 years ago

What is the magnitude of the acceleration of an electron at a point where the electric field has magnitude 6377 n/c and is direc

shusha [124]

Use the magnitude acceleration formula .

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3 years ago

An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is

Lady_Fox [76]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}$

$\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

<h3>☯ <u>By using formula of Lens</u> </h3>

$\\$

$\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}$

$\\$

$\dashrightarrow\:\: \sf{ v = 30 \ cm}$

$\\$

<h3>☯ <u>Now, Finding the magnification </u></h3>

$\\$

$\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}$

$\\$

$\dashrightarrow\:\: \sf{m = -2}$

$\\$

<h3>☯ <u>Hence</u>, $\\$ </h3>

$\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}$

$\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}$

3 0

2 years ago