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2 years ago

Look at the circuit diagram. Which of these components is part of the circuit?

Physics

1 answer:

tangare [24]2 years ago

4 0

I think the answer is c AC power source
Hope this help you?

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Which statement describes a characteristic of an experimental design that

Anarel [89]

Answer:

Explanation:

4 0

3 years ago

Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl

BARSIC [14]

<span>b. The coefficient of static friction for all contacting surfaces is Î¼s=0.35. neglect friction at the rollers.

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6 0

3 years ago

a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?

devlian [24]

For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s² So the average acceleration is 1.75 m/s²

5 0

3 years ago

An oscillator consists of a block attached to a spring (k = 427 N/m). At some time t, the position (measured from the system's e

White raven [17]

Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

$x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)$

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

$\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}$

with this value you can compute the frequency:

$f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz$

the mass of the block is given by the formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg$

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

$\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s$

Finally, the amplitude is:

$x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m$

5 0

2 years ago

Please help and I don't mean to sound rude but, ONLY ANSWER IF YOUR GOING TO DO ALL 4 QUESTIONS

anzhelika [568]

Answer:

for first question is 2

for second question 1

for third question 2

for forth question 1

Explanation:

i hope i helped

6 0

2 years ago