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DedPeter [7]
3 years ago
15

A tightly sealed glass jar is an example of which type of system?

Physics
2 answers:
Katarina [22]3 years ago
7 0
Thermal expansion and contraction, specifically the different coefficients of thermal expansion between glass & metal is a closed system ; D
noname [10]3 years ago
4 0

D. hope it helps. :D

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What is this question
lara31 [8.8K]

Answer:

Its the last one

Explanation:

It is the last one

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3 years ago
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Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an
disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = \frac{1}{2} *m*vf^{2}  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0
4 years ago
Please help asap. I'm terrible with atoms
11Alexandr11 [23.1K]
<span>Correct answer: more, smaller</span>
6 0
3 years ago
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Two sound waves of equal amplitude interference so that compression of one wave falls on the rarefaction of the other. Which sta
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<span>A. No sound is heard 
A*sin(wt)+(-A)*sin(wt)=0 - no sound

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3 0
3 years ago
A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2
Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

Q= kA\frac{dT}{dx}

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C

Then, to find the rate of heat flow per square meter, we have:

\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}

T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness

7 0
3 years ago
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