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3 years ago

PLEASE HELP! Need this ASAP (in 10 minutes)

2 answers:

6 0

The answer to your question here is A,

(I am dragging a very helpful piece of information from a website call “The Physics Classroom” for an explain took by the way so go check them out if you are still having trouble.)

From the paragraph I read, I could pull out “The data convincingly show that wave frequency does not affect wave speed.”

Thus concluding my answer of why the speed is unchanged.

9966 [12]3 years ago

4 0

I think it’s B

step-by-step explanation:

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What is d formula for calculating mechanical power??

lakkis [162]

Answer:

Power = Watt ÷ Time

Explanation:

6 0

3 years ago

Read 2 more answers

You decide to push on a brick wall with all your might for 5 minutes. You

Artist 52 [7]

Answer:

Explanation:W=F*S

W=work =?

F=force =500N

S=displacement =0m

please feel free to ask if you have any questions

5 0

4 years ago

The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com

Flauer [41]

Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

$Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\$

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

$Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\$

<u>Direction = 65.48°</u>

4 0

3 years ago

A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is

vova2212 [387]

solution:

$consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\$ $270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm$

7 0

4 years ago

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

nadezda [96]

Answer:

in the downward movement of the movement when the constant is lost

Explanation:

When the coin is on the piston it has a relationship given by

a = d²x / dt²

the piston position is

x = A cos wt

a = - A w² cos wt

the maximum acceleration is

a = - A w²

When the piston raises the acceleration of gravity and that of the piston go in the same direction, when the piston descends they relate it is contrary to gravity, therefore when the frequency increases, the point where the acceleration of the piston is greater than gravity arrives and the coin loses contact.

The point where you lose contact is

a = g

g = A w²

In short, in the downward movement of the movement when the constant is lost

7 0

3 years ago