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lara31 [8.8K]
2 years ago
15

the autor mentions that shine made several mistakes in her experiment. describe one Major mistake that shine made. support your

description whit details from the text​
Physics
1 answer:
Bond [772]2 years ago
3 0

The author mentions that Shine made several mistakes in her experiment. Describe one major mistake that Shine made.

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A man pushes on a wall what does Newton third law say must happen
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If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.
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3 years ago
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17. Matthew has a piece of wood with mass of 278 g and a volume of 375 cm^3. What is the wood’s density? a. Step 1: Formula ____
Keith_Richards [23]

Answer:

Density =0.74 g/cm^{3}

Explanation:

Step 1: Formula

 Density=\frac{Mass}{Volume}

Step 2: Data

m=278 g

v=375 cm^{3}

Step 3:Solve

d= \frac{278}{375} g/cm^{3}

d=0.74 g/cm^{3}

8 0
2 years ago
Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

4 0
3 years ago
A sample of carbon dioxide (C°p,m = 37.11 J K−1 mol−1) of mass 2.80 g at 27°C is allowed to expand reversibly and adiabatically
My name is Ann [436]

Answer:

182 to 3 s.f

Explanation:

Workdone for an adiabatic process is given as

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats. For carbon dioxide, γ = 1.28

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

K = P₁V₁ʸ

We need to calculate the P₁ using ideal gas equation

P₁V₁ = mRT₁

P₁ = (mRT₁/V₁)

m = 2.80 g = 0.0028 kg

R = 188.92 J/kg.K

T₁ = 27°C = 300 K

V₁ = 500 cm³ = 0.0005 m³

P₁ = (0.0028)(188.92)(300)/0.0005

P₁ = 317385.6 Pa

K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

V₁ = 0.0005 m³

V₂ = 2.10 dm³ = 0.002 m³

1 - γ = 1 - 1.28 = - 0.28

W =

18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)

W = -67.47 (5.698 - 8.4)

W = 182.3 = 182 to 3 s.f

7 0
2 years ago
How is it possible for a single sedimentary rock to contain rock particles animal shells and minerals that crystallized from wat
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