2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!

8 0

2 years ago

An air conditioner operating at steady state maintains a dwelling at 70oF on a day when the outside temperature is 90oF. If the

torisob [31]

30000 btuh /3413 btuh/kw. = 8.8 kw

8.8 kw/.746 kw/hp = 11.8 hp if COP is 1

11.8/3 hp (COP coefficient of performance) = 3.99 COP

>>>So yes a 3.0 hp compressor with a nominal COP of 4 will handle the 30,000 btuh load.

3.2 to 4.5 is expected COP range for an air cooled heat pump or a/c unit.

8 0

2 years ago

If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?

ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: $(m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})$

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

$(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})$