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antiseptic1488 [7]
3 years ago
9

What would be the wavelength of the sound

Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer: 0,96 m

Explanation: speed in concrete is 4300 m/s

According to wave equation speed v = lambda × frequency

Lambda = speed /frequency

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A meterstick is placed on a pivot point of 42.5cm and a 45g mass is hung at the 20cm mark. When released the meterstick remains
Vikki [24]
Hey i dont have an answer but i need the points for finals today. Thank you
6 0
3 years ago
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A 0.5 kg stone is raised from 1m to 2m height from the ground. what is the change in potential energy of the stone?
Usimov [2.4K]

Given: The mass of stone (m) = 0.5 kg

Raised from heights (h₁) = 1.0 m to (h₂) = 2.0 m

Acceleration due to gravity (g) = 9.8 m/s²

To find: The change in potential energy of the stone

Formula: The potential energy (P) = mgh

where, all alphabets are in their usual meanings.

Now, we shall calculate the change in potential energy of the stone

Δ P = P₂ - P₁ = mg (h₂ - h₁)

or,                = 0.5 kg ×9.8 m/s² ×(2.0 m - 1.0 m)

or,                = 4.9 J

Hence, the required change in the potential energy of the stone will be 4.9 J

4 0
3 years ago
the ratio of the energy per second radiated by the filament of a lamp at 250k to that radiated at 2000k, assuming the filament i
Naily [24]

Answer:

(a) \frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b) P =  0.816 Watt

Explanation:

(a)

The power radiated from a black body is given by Stefan Boltzman Law:

P = \sigma AT^4

where,

P = Energy Radiated per Second = ?

σ = stefan boltzman constant = 5.67 x 10⁻⁸ W/m².K⁴

T = Absolute Temperature

So the ratio of power at 250 K to the power at 2000 K is given as:

\frac{P_{250k}}{P_{2000k}}=\frac{\sigma A(250)^4}{\sigma A(2000)^4}\\\\\frac{P_{250k}}{P_{2000k}}=2.4\ x\ 10^{-4}

(b)

Now, for 90% radiator blackbody at 2000 K:

P = (0.9)(5.67\ x\ 10^{-8}\ W/m^2.K^4)(1\ x\ 10^{-6}\ m^2)(2000\ K)^4

<u>P =  0.816 Watt</u>

7 0
3 years ago
Help pls i need this right now
pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

5 0
3 years ago
The moment of water from ocean through the atmosphere and back
melisa1 [442]

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There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

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3 years ago
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