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Irina-Kira [14]
4 years ago
10

How do you know if an experimental result is acceptable or trustworthy? What gives you confidence that your data are trustworthy

?
Physics
1 answer:
Virty [35]4 years ago
6 0

Explanation:

For an experimental result to be considered acceptable, all relevant variables involved in the experiment must be taken into account, by isolating it, performing it under controlled conditions and modifying the conditions under which it takes place. This, with the objective of excluding alternative explanations in the analisis of the experimental data. Therefore, if these steps are followed appropriately, experimental data are trustworthy. The reliability of the experiment increases when it is replicated by other researchers and the same results are obtained.

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hovering mosquito is hit by a raindrop that is 45 times as massive and falling at 8.1 m/s , a typical raindrop speed. How fast i
Y_Kistochka [10]

Answer:

7.92 m/s

Explanation:

m_1 = Mass of raindrop = 45m_2

m_2 = Mass of mosquito

u_1 = Initial Velocity of raindrop = 8.1 m/s

u_2 = Initial Velocity of mosquito = 0 m/s

v = Velocity of center of mass

For elastic collision

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{45m_2\times 8.1 + m_2\times 0}{45m_2 + m_2}\\\Rightarrow v=\frac{364.5m_2}{46m_2}\\\Rightarrow v=7.92\ m/s

Hence, the velocity of the attached mosquito, falling immediately afterward is 7.92 m/s

4 0
3 years ago
Help me pls its for my science class i need to show my work
Lostsunrise [7]

Answer:

P = 5880  J

Explanation:

Given that,

The mass of the block, m = 30 kg

The block is sitting at a height of 20 m.

The block will have gravitational potential energy. The formula for gravitational potential energy is given by :

P=mgh\\\\=30\times 9.8\times 20\\\\=5880\ J

So, the required potential energy is equal to 5880  J.

6 0
3 years ago
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Delicious77 [7]
THE answer the table is holding the book up would be a normal force 
6 0
4 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
4 years ago
Please help!!! what is the main point of paragraph 3?
Nikitich [7]
There’s no picture:(
8 0
3 years ago
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