Ask question

Ask question

All categories

3 years ago

5

The velocity of sound apparatus is used in an investigation to determine the frequency of an unknown tuning fork. The temperatur

e of the room is 25° Celsius, the first antinode is at .75meters with the second position at 2.25.
A. Determine the wavelength of the tuning fork.

B. Determine the speed of sound in that room.

C. Determine the frequency of the tuning fork.

1 answer:

Hitman42 [59]3 years ago

6 0

Answer:

Explanation:

The relationship of the speed of sound, its frequency, and wavelength is the same as for all waves: vw = fλ, where vw is the speed of sound, f is its frequency, and λ is its wavelength.

You might be interested in

In a 50 km/h head-on crash test, the steering column of passenger car 1 moved 3 cm upward and 2 cm rearward. The steering column

Softa [21]

Answer:

Car 1

Explanation:

The steering column which moves the least is less likely to to the driver's chest ordinarily. Driver tends to remain in motion until restrained. Assuming a seat belt not airbag

Generally one would compute a vector find direction and distance. This is like solving for a hypotenuse / in a right angled triangle problem. On face value the column moving the least is safer. The 6/24 would hit the upper chest, face, or possibly break the neck.

hence, car 1 moved 3 cm upward and 2 cm rearward is safer.

4 0

3 years ago

A force of 1000 newtons was necessary to lift a rock. A total of 3000 joules of work was done. How far was the rock lifted?

CaHeK987 [17]

Answer:

Explanation:

3 meters

4 0

2 years ago

Can spoon ring like a bell prove it

mote1985 [20]

Yes spoon can sound like a bell. To prove this, we perform an experiment.The handle of the spoon is tied at the mid point of the string, then wrap the ends of the string around pointer fingers. Now place fingers in ears. Lean over so that spoon hangs freely and swing the spoon so it taps against a door.
A sound is produced because the spoon vibrated, causing sound waves to travel up the string and into ears.

4 0

3 years ago

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

Katyanochek1 [597]

Answer:

The force is $F = 1041.7N$

Explanation:

The moment of Inertia I is mathematically evaluated as

$I = MR_A^2$

Substituting $1.9kg$ for M(Mass of the wheel) and $\frac{66cm}{2} * \frac{1m}{100cm} = 0.33m$ for $R_A$ (Radius of wheel)

$I = 1.9 * 0.33^2$

$= 0.207kgm^2$

The torque on the wheel due to net force is mathematically represented as

$\tau = FR_B - F_rR_A$

Substituting 135 N for $F_r$ (Force acting on sprocket), $\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m$ for $R_B$ (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

$\tau = F (0.0435) - 135 (0.33)$

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of $\alpha = 3.70 rad/s^2$ and this torque can also be represented mathematically as

$\tau = \alpha I$

Now equating the two equation for torque

$F (0.0435) - 135 (0.33) = \alpha I$

Making F the subject

$F = \frac{\alpha I + (135*0.33) }{0.0435}$

Substituting values

$F = \frac{(3.70 * 0.207) + (135*0.33)}{0.0435}$

$= 1041.7N$

4 0

3 years ago

A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus

babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

$I = m\cdot (\vec{v}_{2} - \vec{v_{1}})$ (1)

Where:

$I$ - Impulse, in kilogram-meters per second.

$m$ - Mass, in kilograms.

$\vec{v_{1}}$ - Initial velocity of the hockey park, in meters per second.

$\vec{v_{2}}$ - Final velocity of the hockey park, in meters per second.

If we know that $m = 0.2\,kg$ , $\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]$ and $\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]$ , then the impulse applied by the stick to the park is approximately:

$I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]$

$I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]$

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.

8 0

3 years ago