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sergij07 [2.7K]
3 years ago
9

A plane progressive wave is represented by the equation y =2sin (2000πt- 0.5x) the symbols have their usual meanings. what is th

e frequency of the wave
Physics
1 answer:
frutty [35]3 years ago
5 0

The frequency of the wave is the frequency of that sine function ... 2,000π ... but that's the <u>angular </u>frequency, in radians per second.  

To convert angular frequency to Hz, we have to divide by 2π.

So the frequency of this wave is <em>1,000 Hz. </em>

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I need this right ASAP!!! plz plz plz
UkoKoshka [18]

Answer:

The answer is D

Nucleus is neutral

8 0
3 years ago
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A 403.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 320.0 K. How much heat must the smelter p
aleksklad [387]
The answer is 2.49 x 10^5 KJ. This was obtained (1) use the formula for specific heat to achieve Q or heat then (2) get the energy to melt the copper lastly (3) Subtract both work and the total energy required to completely melt the copper bar is achieved.
7 0
3 years ago
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A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.
densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

KE = \frac{1}{2} mv^2

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (5*10^{-3})(300)^2

KE =  225J

On the other hand the required Energy to heat up t melting point is

Q_1 = mC_p \Delta T

Q_2 = L_f m

Where,

m = Mass

C_p =Specific Heat

\Delta T =Change at temperature

L_f = Latent heat of fussion

Heat required to heat up to melting point,

Q = Q_1+Q_2

Q = mC_p \Delta T+L_f m

Q = 5*0.128*(327-20) + 5*24.7

Q = 310J

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0
3 years ago
When you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Ea
Klio2033 [76]

Answer:

5.66 × 10⁻²³ m/s

Explanation:

If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.

Since initial momentum = final momentum,

mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.

My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?

So mv₁ + M × 0 = m × 0 + MV₂

mv₁ = MV₂

V₂ = mv₁/M =  54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s

5 0
3 years ago
In the equation for the gravitational force between two objects, what does G represent? the gravitational constant the gravitati
kumpel [21]
G is the gravitational constant, which is approximately 6.6x10^-11 Nm/s^2. It has the same value regardless of the masses of both objects or the distance between them.
3 0
2 years ago
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