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GaryK [48]
3 years ago
15

Jerry is conducting an experiment to test friction. His setup is shown below. If he wants to test the friction between different

types of materials, what should he change in the setup above? A. Replace the mass with a different size mass B. Replace the string with a different length string C. Replace the spring scale with a different kind of scale D. Replace the wooden block with a different kind of block

Physics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

The correct option is D

Explanation:

This question is incomplete because of the absence of the setup which as been attached below. The setup shows/determines/tests the friction of wood (which is a block material), since Jerry wants to test the friction between different types of materials, he will have to replace the wooden block with another type of block material of choice so as to determine the friction of that also.

In order to have a comprehensive experiment, Jerry can use 4-5 different types of block material in the course of the experiment.

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Describe how tree rings indicate time. Do the same for ice cores and varves?
tatyana61 [14]

Answer:

counting tree rings can indicate age

Explanation:

hope i helped at least a little bit :)

7 0
3 years ago
What is the value of x in the equation below 1+2e^x+1=9
GuDViN [60]
<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of x:

1+2e^x+1=9   (1)

Firstly, we have to clear x:

2e^x=9-1-1  

2e^x=7  

e^x=\frac{7}{2}     (2)

Applying<u> Natural Logarithm</u> on both sides of the equation (2):

ln(e^x)=ln(\frac{7}{2})     (3)

xln(e)=ln(\frac{7}{2})     (4)

According to the Natural Logarithm rules xln(e)=x, so (4) can be written as:

x=ln(\frac{7}{2})     (5)

Finally:

x=1.252    

3 0
3 years ago
List the metric base units for the following:
77julia77 [94]
Meter
meter
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4 0
3 years ago
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What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring
mote1985 [20]

Answer:

The orbital velocity v  =  16.4 \ km/s

The period is  T =  14.8 \ hours

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the planet , this is mathematically represented as

       \frac{GM_s  *  m }{r^2 }  = m w^2 r

=>    w = \sqrt{ \frac{GM}{r^3} }

Here G is the gravitational constant with value  G = 6.67*10^{-11}

        M_s  is the mass of with value  M_s  =5.683*10^{26} \  kg

        r is the is distance from the center of the  to the  outer edge of the  A ring

i.e r = R  + D  

Here R  is the radius of the planet   with value  R  = 60300 \ km

         D  is the distance from the  equator to the outer edge of the  A ring with value  D = 80000 \  kg

So  

       r =80000 + 60300

=>    r =140300 \ km  = 1.4*10^{8} \  m

So

    =>    w = \sqrt{ \frac{ 6.67*10^{-11}*  5.683*10^{26}}{[1.4*10^{8}]^3} }

    =>    w =  1.175*10^{-4} \ rad/s

Generally the orbital velocity is mathematically represented as

       v  = w * r

=>     v  = 1.175*10^{-4}   * 1.4*10^{8}

=>     v  = 1.64*10^{4} \  m /s =  16.4 \ km/s

Generally the period is mathematically represented as

     T   =  \frac{2 \pi }{w }

=> T   =  \frac{2 *  3.142  }{ 1.175 *10^{-4} }

=> T   = 53473 \ second = 14.8 \ hours

Answer:

The orbital velocity v  =  16.4 \ km/s

The period is  T =  14.8 \ hours

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational  force between the ring particle and the , this is mathematically represented as

       \frac{GM_s  *  m }{r^2 }  = m w^2 r

3 0
3 years ago
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.

4 0
2 years ago
Read 2 more answers
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