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Tema [17]
3 years ago
12

8. If horizontal velocity is 5 m/s, and vertical velocity is 8 m/s, what is the

Physics
1 answer:
zhenek [66]3 years ago
7 0

Answer:

9.4

Explanation:

magnitude is the sum of the squares.

\sqrt{5^2+8^2} =9.4339\\

If you are given horizontal and vertical components, treat those as the rise and run of a triangle, the rise of 8 with a run of 5 and you want to find the hypotenuse.

How do you find the long side of a triangle?  

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Answer:

There are six kinds of forces that act on objects when they come into contact with one another: Normal force, applied force, frictional force, tension force, spring force and resisting force. These forces make objects change their motion or movement , the act of going from one place to another.

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Which of the following is an example of a lever? knife ramp pencil sharpener wheelbarrow
Kay [80]
I believe it is a knife
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3 years ago
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A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe
sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let \alphabe the angular acceleration

2 T\times r=I\times \alpha

2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha

\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}

\alpha =\frac{3.24\times 10^3}{m} rad/s^2

\omega =\omega _0+\alpha \cdot t

\omega =0+\frac{3.24\times 10^3}{m}\times 0.74

\omega =\frac{2.4\times 10^3}{m} rad/s

Angular momentum

L=I\omega

L=0.5\times mr^2\times \omega

L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}

L=0.01051 kg-m^2/s

4 0
3 years ago
What is the weight of a bag that has the mass of 18.0kg
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Please answer this question sqdancefan​
AleksandrR [38]

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%.

_____

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%.

_____

<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.

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