We know the equation
weight = mass × gravity
To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.
So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)
Therefore,
100N = mass × 10
mass= 100N/10
mass= 10 kg
Now, all we need are the moon's gravitational field strength and to apply this to the equation
weight = 10kg × (gravity on moon)
For this we use general equation for gases. Our variables represent:
p- pressure
v-volume
t- temperature
P1V1/T1 = P2V2/T2
in this equation we know:
P1,V1 and T1, T2 and V2.
We have one equation and 1 unknown variable.
P2 = T2P1V1/T1V2 = 1.1atm
Enough data points would have been collected when specific data for upward and downward motion are collected.
<h3>
What is a Data point?</h3>
A data point is a discrete unit of information. This information is always unique.
<h3>Free fall</h3>
An object subjected to free fall is under the influence of gravity. An object subjected has different time of motion for upward and downward motion.
Thus, we can conclude that enough data points would have been collected when specific data for upward and downward motion are collected.
Learn more about free fall here: brainly.com/question/10909077
V=d/t=>v=400m/44.40s=>v=9m/s
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC.
Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules
Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used
q=mCΔT
11466.58 Joules = 65.1g x 4.18 J / g C x ΔT
11466.58/(65.1gx4.18)=ΔT
ΔT=42.14oC
So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>