If you are asked to determine an objects speed, what information must you have?

A bugle can be thought of as an open pipe. If a bugle were straightened out, it would be 2.65 mlong.a.) If the speed of sound is

Nezavi [6.7K]

Answer:

(a). The lowest frequency is 64.7 Hz.

(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.

Explanation:

Given that,

Length = 2.65 m

Speed of sound = 343 m

We need to calculate the wavelength

Using formula of wavelength

$\lambda=2l$

Put the value into the formula

$\lambda=2\times2.65$

$\lambda=5.3\ m$

(a). We need to calculate the lowest frequency

Using formula of frequency

$f_{1}=\dfrac{v}{\lambda_{1}}$

Put the value into the formula

$f_{1}=\dfrac{343}{5.3}$

$f_{1}=64.7\ Hz$

(b). We need to calculate the next two resonant frequencies for the bugle

Using formula of resonant frequency

$f_{2}=\dfrac{v}{\lambda_{2}}$

$f_{2}=\dfrac{v}{l}$

Put the value into the formula

$f_{2}=\dfrac{343}{2.65}$

$f_{2}=129.4\ Hz$

For third frequency,

$f_{3}=\dfrac{v}{\lambda_{3}}$

$f_{3}=\dfrac{3v}{2l}$

Put the value into the formula

$f_{3}=\dfrac{3\times343}{2\times2.65}$

$f_{3}=194.2\ Hz$

Hence, (a). The lowest frequency is 64.7 Hz.

(b). The next two resonant frequencies for the bugle are 129.4 Hz and 194.2 hz.

3 0

3 years ago

How long does it take an automobile traveling in the left lane of a highway at 60.0 km/h to overtake (become even with) another

pashok25 [27]

Answer:

4.95 seconds

Explanation:

It is given that the speed of the cars are constant and not accelerating

The relative speed between the two cars will be the difference in their speeds

$S_r=60-20\\\Rightarrow S_r=40\ km/h$

Converting to m/s

$1\ km/h=\frac{1}{3.6}\ m/s$

$\\\Rightarrow 40\ km/h=40\times \frac{1}{3.6}\ m/s\\ =\frac{100}{9}\ m/s\\ =11.11\ m/s$

Time = Distance / Speed

$Time=\frac{55}{\frac{100}{9}}=4.95\ seconds$

Time it would take the faster car to cross the slower car is 4.95 seconds

8 0

3 years ago

A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity

WARRIOR [948]

Answer:

The charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg

Explanation:

From the formulae

F = qvB and F = mv²/r

Where F is Force

q is charge

v is speed

B is magnetic field strength

m is mass

and r is radius

Then,

qvB = mv²/r

qB = mv/r

We can write that

q/m = v/rB ---- (1)

Also

From Electric force formula

F = Eq

Where E is the electric field

and magnetic force formula

F = Bqv

Since, electric force = magnetic force

Then, Eq = Bqv

E = Bv

∴ v = E/B

Substitute v = E/B into equation (1)

q/m = (E/B)/rB

∴ q/m = E/rB²

(NOTE: q/m is the charge to mass ratio)

From the question,

E = 3.10 ×10³ N/C

r = 4.20 cm = 0.0420 m

B = 0.360 T

Hence,

q/m = 3.10 ×10³ / 0.0420 × (0.360)²

q/m = 569517.9306 C/kg

q/m = 5.7 × 10⁵ C/kg

Hence, the charge-to-mass ratio of the particle is 5.7 × 10⁵ C/kg.

7 0

3 years ago

A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from

ivann1987 [24]

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

$m$ - Mass of the object, in kilograms.

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$ , $M = 1.5\,kg$ , $m = 0.54\,kg$ and $r = 0.1\,m$ , then the angular speed of the turntable after the event is: