Answer:
-194 m/s
Explanation:
We are taking as a reference point the position of the sun
In this case,
The spacecraft is moving with a velocity of +126 m/s
The asteroid is moving past in a backwards direction at 68 m/s relative to the spacecraft.
Speed of the asteroid - speed of the spacecraft = 68m/s
Speed of the asteroid = 68 m/s + 126 m/s
Speed of the asteroid = 194 m/s (backwards)
Beacuse the asteroid is moving in the opposite direction of the spacecraft
The asteroid has a velocity of -194m/s relative to the position of the Sun
Im pretty sure the answer would be A
Given the mass of R-134a m = 300kg; Volume of the container V = 9 cu. meter; Temperature of R-134a T = 10 degrees Celsius;
Formula of specific volume : v = V / m = 9 / 300 = 0.03 cu. m / kg.
At T = 10 degrees Celsius from saturated R-134a tables, vf = 0.0007930 cu. m /kg; vg = 0.049403 cu. m/kg. We know v = vf + x (vg - vf), so 0.03 = 0.0007930 + x (0.049403 - 0.0007930), which makes x = 0.601.
Specific enthalpy of R-134a in the container is h = hf + x*hfg = 65.43 + (0.601 * 190.73). Answer is 180.0587 kJ/kg
I also agree to A. The force will always pull back a lot deals with gravitational pull
Answer:
<h2>A. twice</h2>
Explanation:
Spring potential Energy is expressed as E = 1/2 ke²
k = spring constant
e = displacement of the string
If Spring A has a spring constant of 100 N/m and displacement of 0.2m, the spring potential energy will be;
= 1/2 * 100 * 0.2²
= 50 * 0.04
= 2Joules
If spring B has a spring constant of 200 N/m and displacement of 20m, the spring potential energy will be;
= 1/2 * 200 * 0.2²
= 100 * 0.04
= 4Joules
From the values gotten, it can be seen that the spring potential energy of spring B is twice that of A