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3 years ago

Physics multiple choice

Physics

2 answers:

Aleksandr-060686 [28]3 years ago

3 0

10. b

11.d

12. b

13.e

I hope this helps

Kobotan [32]3 years ago

3 0

Answer:

10. a

11.d

12. b

13.e

Explanation:

I hope I can help you have a great weekend

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To practice Problem-Solving Strategy 16.1 Standing waves. An air-filled pipe is found to have successive harmonics at 800 HzHz ,

bazaltina [42]

Answer:

Length of the pipe = 53.125 cm

Explanation:

given data

harmonic frequency f1 = 800 Hz

harmonic frequency f2 = 1120 Hz

harmonic frequency f3 = 1440 Hz

solution

first we get here fundamental frequency that is express as

2F = f2 - f1 ...............1

put here value

2F = 1120 - 800

F = 160 Hz

and

Wavelength is express as

Wavelength = Speed ÷ Fundamental frequency ................2

here speed of waves in air = 340 m/s

so put here value

Wavelength =340 ÷ 160

Wavelength = 2.125 m

Length of the pipe will be

Length of the pipe = 0.25 × wavelength ......................3

put here value

Length of the pipe = 0.25 × 2.125

Length of the pipe = 0.53125 m

Length of the pipe = 53.125 cm

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3 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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3 years ago

In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

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3 years ago