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Kazeer [188]
3 years ago
5

3.00 m^3 of water is at 20.0°C.

Physics
1 answer:
romanna [79]3 years ago
7 0

Answer:

\triangle V = 0.02484m^3

Explanation:

Given

V_1 = 3.00m^3 --- initial volume

T_1 = 20.0^oC --- initial temperature

T_2 = 60.0^oC --- final temperature

\gamma = 207*10^{-6 ---  coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

\triangle V = \gamma * V_2 * (T_2 - T_1)

So, we have:

\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)

\triangle V = 207 * 10^{-6} * 3.00 * 40.0

\triangle V = 0.02484m^3

The volume will expand by 0.02484m^3

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Answer:

630 Hz.

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As we are considering the one end open pipe. So for the sound wave there will be a pressure node at the open end of the tube as at that place the molecules can not move back and forth. However on the closed end there will be a flow node as the water molecules their are moving back and forth. So it will produces the resonance at the positions 1/4, 3/4.......

we can find the wavelength by multiplying the levels distance by 2.

λ = 2 × 0.27 m = 0.54

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4 years ago
What is the speed of a wave with a wavelength of 10 mm and a frequency of 5.0 Hz?
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3 years ago
A wooden block with mass 1.45 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 29.0 deg
Yakvenalex [24]

Answer:

76.3 J

Explanation:

I'm assuming the distance of 4.60 m is along the incline, not the vertical distance from the bottom.  I'll call this distance d, so h = d sin θ.

Initial energy = final energy

Energy in spring = gravitational energy + kinetic energy + work by friction

E = mgh + 1/2 mv² + Fd

We need to find the force of friction.  To do that, draw a free body diagram.

Normal to the incline, we have the normal force pointing up and the normal component of weight (mg cos θ).

Sum of the forces in the normal direction:

∑F = ma

N - mg cos θ = 0

N = mg cos θ

Friction is defined as:

F = Nμ

Plugging in the expression for N:

F = mgμ cos θ

Substituting:

E = mgh + 1/2 mv² + (mgμ cos θ) d

E = mg (d sin θ) + 1/2 mv² + (mgμ cos θ) d

E = mgd (sin θ + μ cos θ) + 1/2 mv²

Given:

m = 1.45 kg

g = 9.90 m/s²

d = 4.60 m

θ = 29.0°

μ = 0.45

v = 5.10 m/s

Solving:

E = mgd (sin θ + μ cos θ) + 1/2 mv²

E = (1.45) (9.80) (4.60) (sin 29.0 + 0.45 cos 29.0) + 1/2 (1.45) (5.10)²

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