All categories

3 years ago

The musical tone a piano has a frequency of 346HZ and a wavelength of 1.3 M what is the speed of the sound

Physics

1 answer:

prohojiy [21]3 years ago

5 0

Answer:

v= 449.8 m/s

Explanation:

Given data

Frequency= 346Hz

Wave length= 1.4m

The expression below is used to find the speed

$v= f \lambda\\\\$

substitute

$v= 346*1.3\\\\v= 449.8 m/s$

Hence the speed is v= 449.8 m/s

You might be interested in

Invader Zim’s spaceship is sitting at rest in outer space. The ship then accelerates at a uniform rate of 12 m/s2 for 10 seconds

melisa1 [442]

Answer:

120 m/s

Explanation:

Given:

v₀ = 0 m/s

a = 12 m/s²

t = 10 s

Find: v

v = at + v₀

v = (12 m/s²) (10 s) + 0 m/s

v = 120 m/s

6 0

3 years ago

Paul lifts a sack weighing 245 newtons vertically from the ground and places it on a platform at a height of 0.7 meters. If he t

d1i1m1o1n [39]

Answer:

B. 17.15 watts

Explanation:

Given that

Time = 10 seconds

height = distance = 0.7 meters

weight of sack = mg = F = 245 newtons

Power = work done/ time taken

Where work done = force × distance

Substituting the given parameters into the formula

Work done = 245 newton × 0.7 meters

Work done = 171.5 J

Recall,

Power = work done/time

Power = 171.5 J ÷ 10

Power = 17.15 watts

Hence the power expended is B. 17.15 watts

6 0

3 years ago

Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke

umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

Thrust = $v_{e} \frac{dM}{dt}$

$v_{f}$ -v₀ = v_{e} $ln ( \frac{M_{o} }{M_{f}} )$

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

The final mass is the mass of the engines + the mass of the rocket

M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

thrust = 5.26 N

t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

thrust = v_{e} $\frac{M_{f} - M_{o} }{t_{f} - t_{o} }$

v_{e} = thrust $\frac{\Delta t}{\Delta M}$

v_{e} = 5.26 $\frac{1.90}{0.080 -0.0927}$

v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

v_{f}-0 = v_{e} $ln ( \frac{M_{o} }{M_{f} } )$

we calculate

v_{f} = 786.93 ln (0.0927 / 0.080)

v_{f} = 115.95 m / s

5 0

3 years ago

A hot air balloon must be designed to support a basket, cords, and one person for a total payload weight of 1300 N plus the addi

RSB [31]

Answer:

r = 4.44 m

Explanation:

For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid

B = ρ g V

Now let's use Newton's equilibrium relationship

B - W = 0

B = W

The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)

σ = W / A

W = σ A

The area of a sphere is

A = 4π r²

W = W₁ + σ 4π r²

The volume of a sphere is

V = 4/3 π r³

Let's replace

ρ g 4/3 π r³ = W₁ + σ 4π r²

If we use the ideal gas equation

P V = n RT

P = ρ RT

ρ = P / RT

P / RT g 4/3 π r³ - σ 4 π r² = W₁

r² 4π (P/3RT r - σ) = W₁

Let's replace the values

r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000

r² (11.81 r -0.060) = 13000 / 4pi

r² (11.81 r - 0.060) = 1034.51

As the independent term is very small we can despise it, to find the solution

r = 4.44 m

3 0

4 years ago

A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a

VashaNatasha [74]

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

$r_1 = 45 sin15 \hat i + 45 cos15 \hat j$

$r_1 = 11.64 \hat i + 43.46\hat j$

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

$r_2 = 30 cos15\hat i + 30 sin15 \hat j$

$r_2 = 28.98\hat i + 7.76 \hat j$

now the displacement of the boat is given as

$d = r_2 - r_1$

$d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)$

$d = 17.34 \hat i - 35.7 \hat j$

so the magnitude is given as

$d = \sqrt{17.34^2 + 35.7^2}$

$d = 39.7 km$

4 0

3 years ago

The musical tone a piano has a frequency of 346HZ and a wavelength of 1.3 M what is the speed of the sound￼

The musical tone a piano has a frequency of 346HZ and a wavelength of 1.3 M what is the speed of the sound