Answer:
10250 N/C leftwards
Explanation:
QA = 4 micro Coulomb
QB = - 5 micro Coulomb
AP = 6 m
BP = 2 m
A is origin, B is at 4 m and P is at 6 m .
The electric field due to charge QA at P is EA rightwards

The electric field due to charge QB at P is EB leftwards

The resultant electric field at P due the charges is given by
E = EB - EA
E = 11250 - 1000 = 10250 N/C leftwards
Answer:
B)a tool to drop temperatures, mercury, an electric current, and a tool to measure resistance
Answer:
The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet
Explanation:
The given parameters are;
The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M
The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R
The weight of the cadet on Earth = 800 N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.
Answer:
the answer is the temperatures of both beakers' water will increase by the same amount...
Explanation:
I know this because i just did it on study island