Roughly 13.8 billion years old according to science
E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

hence, the speed of the Hubble is approximately 384km/min
Answer: independent variable: Size of the feather.
Explanation:
In an experiment, the manipulated/independent variable is, as the name implies, the variable that the scientist can control.
In this case, the scientist has only one variable that he can control at will, and this is the size of the feather (he can choose which feather he uses for the experiment)
So the manipulated variable will be the size of the feather.
And the dependent variable is the one that "answers" to the changes in the manipulated variable.
In this case, will be the time that it takes to the feather to fall to the ground.
Answer:
The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω
Explanation:
Adding Resistances in serie is the way to build de largest equivalent value possible.
Rt= R1+R2+R3
Rt= 6.32 + 8.13 + 2.29
Rt= 16.74Ω