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inn [45]
2 years ago
14

Landon attends an early childhood program that is located at a community center which also runs an adult care program.

Physics
1 answer:
Ira Lisetskai [31]2 years ago
3 0
Good for Landon. What’s the question?
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Which one of the following accurately describes the force of gravity?
Aloiza [94]
B. The gravity acceleration is in the same direction as the force of gravity, and thus towards the centre of the earth

4 0
3 years ago
What are the two factors that affect the period of a pendulum?
klemol [59]
The two factors that affect the period of a pendulum are the length of the string and the distance in which the pendulum falls.

Hope this helps! I would greatly appreciate a brainliest! :)
3 0
3 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
3 years ago
What is potential difference also called?
malfutka [58]
Voltage - V = IR : v= voltage i= current r= resistance
6 0
2 years ago
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column
scoray [572]

Answer:

The answer is "1155\ \frac{kg}{m^3}"

Explanation:

Please find the complete question in the attached file.

p = p_0 + ?gh

pi = pressure only at two liquids' devices

PA = pressure atmosphere.

1 = oil density

2 = uncertain fluid density

h_1 = 11 \ cm\\\\h_2= 3 \ cm

The pressures would be proportional to the quantity 11-3 = 8 cm from below the surface at the interface between both the oil and the liquid.

\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\

       = \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}

8 0
3 years ago
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