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miskamm [114]
2 years ago
13

A sound wave in air has a frequency of 354 Hz and travels with a speed of 343 m/s. How far apart are the wave crests in meters.

Physics
1 answer:
dsp732 years ago
6 0

Given :

The frequency of travelling sound = f = 282 Hz

The speed of travelling sound = c = 343 m/s

To Find :

The distance of waves crests

Solution :

∵ wavelength of sound wave =

i.e         =

Or,        =

∴          = 1.12   m

So, The wavelength of sound wave =   = 1.21  m

Thus , The distance of waves crests =wavelength of sound wave = 1.21 m

Hence, The distance of waves crests is 1.21 meters    Answer

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A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a ra
worty [1.4K]

Answer:

The torque on the pulley, when the system is motionless is approximately 9.81 N·m

Explanation:

The given parameters are;

The mass of the object = 2 kg

The friction between the rope and the pulley = 0

The mass of the rope, m_r = 0.5 kg

The mass of the pulley, m_p = 0.01 kg

The radius of the pulley, r = 0.25 m

The torque on the pulley, τ = I·α = F × D

The torque on the pulley, when the system is motionless, τ = F × D

Where;

F = The force acting on the pulley rope = The weight of the mass ≈ 2 kg × 9.81 m/s² = 19.62 N

D = The diameter of the pulley = 2×r = 2 × 0.25 m = 0.5 m

Therefore;

τ = 19.62 N × 0.5 m = 9.81 N·m

The torque on the pulley, when the system is motionless, τ ≈ 9.81 N·m.

4 0
2 years ago
A pulley is able to lift a mass of 25 kg 0.30 m with an applied force of 50 N over a distance of 1.5 m. What is the ideal mechan
spin [16.1K]

The ideal mechanical advantage of the pulley system is 3

7 0
3 years ago
A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2
Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel I=25kgm^2

Initial angular velocity of the wheel \omega _0=10rad/sec

Angular acceleration \alpha =15rad/sec^2

(a) We know that \omega =\omega _0+\alpha t

We have given t = 2 sec

So \omega =10+15\times  2=40rad/sec

Now \Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad

(b) After 3 sec \omega =10+15\times 3=55rad/sec

We know that kinetic energy is given by Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J

7 0
3 years ago
When are all the forces acting upon an object balanced?
Rudik [331]

Answer:

When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.

Explanation:

7 0
2 years ago
A rocket is fired at 100 m/s at an angle of 37, how many seconds is it in the air?
Sonbull [250]

Answer:

12.3 seconds

Explanation:

time in air = 2(v)/g

2(100sin(37))/9.8

=12.28 seconds

7 0
3 years ago
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