The answer would be that they are close to water hope this helps!
<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>
Answer:
The answer to your question is: 15 m/s2
Explanation:
Equation x = at3 - bt2 + ct
a = 4.1 m/s3
b = 2.2 m/s2
c = 1.7 m/s
First we find x at t = 4.1 s
x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)
x = 4.1(68.921) - 2.2(16.81) + 6.97
x = 282.58 - 36.98 + 6.98
x = 252.58 m
Now we find speed
v = x/t = 252.58/ 4.1 = 61.6 m/s
Finally
acceleration = v/t = 61.6/4.1 = 15 m/s2
Answer:
7 units, north
Explanation:
This is a problem of vector subtraction. We have:
- Vector A: magnitude 2 units, direction to the north
- Vector B: magnitude 5 units, direction to the south
If we take the north as positive direction, we can write
Since we want to find 
(vector subtraction), we have to change the sign of B, so we find:
And the positive sign means the direction is north.
