The way I actually did that it was just like a little bit of a panic attack and I was like literally dying laughing at my chrome book mark and I was like literally dying laughing at the park I was laughing so loud and I’m literally gonna laughing so I can’t do tell him what he says I don’t think I
Answer:
(d) a net external force must be acting on the system
Explanation:
Momentum is given as the product of mass and velocity.
P = MV
According to Newton's second law of motion, " Force applied to a body (system) is directly proportional to the rate of change of momentum of the body (system) which takes place in the direction of the applied force (external force).
F ∝ΔMV
Therefore, If the total momentum of a system is changing, a net external force must be acting on the system.
(d) a net external force must be acting on the system
160w/4 = 40w
1000J/ 4= 250J
a 40w light bulb will consume 250J
Answer:
1st t=d/s
2nd s=d/t
5th d=st
are all of the variations I came up with
Explanation:
the original formula for speed is s=d/t
then, I created a ratio of s/1=d/t and cross multiplied to find d=st
then, I isolated the t in one side by dividing by s on both sides.
Answer:
(D) 4
Explanation:
The percentage error in each of the contributors to the calculation is 1%. The maximum error in the calculation is approximately the sum of the errors of each contributor, multiplied by the number of times it is a factor in the calculation.
density = mass/volume
density = mass/(π(radius^2)(length))
So, mass and length are each a factor once, and radius is a factor twice. Then the total percentage error is approximately 1% +1% +2×1% = 4%.
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If you look at the maximum and minimum density, you find they are ...
{0.0611718, 0.0662668} g/(mm²·cm)
The ratio of the maximum value to the mean of these values is about 1.03998. So, the maximum is 3.998% higher than the "nominal" density.
The error is about 4%.
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<em>Additional comment</em>
If you work through the details of the math, you will see that the above-described sum of error percentages is <em>just an approximation</em>. If you need a more exact error estimate, it is best to work with the ranges of the numbers involved, and/or their distributions.
Using numbers with uniformly distributed errors will give different results than with normally distributed errors. When such distributions are involved, you need to carefully define what you mean by a maximum error. (By definition, normal distributions extend to infinity in both directions.) While the central limit theorem tends to apply, the actual shape of the error distribution may not be precisely normal.