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Zarrin [17]
3 years ago
15

Which of the following materials could have been found in the giant cloud that formed the solar system?

Physics
2 answers:
iren2701 [21]3 years ago
6 0
Liquid and solid water were not in the giant gas cloudr
Brilliant_brown [7]3 years ago
4 0
3. Carbon dioxide and oxygen

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A party shop delivers helium-filled balloons to homes and businesses. The owners realize from experience that on hot summer days
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Explanation:

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I can't seem to get the right angular acceleration and also not sure how to do part b. Help will be much appreciated.
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Answer:It’s 5 I believe

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3 years ago
A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
pogonyaev

Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

Because tension is equal to centripetal force

Substitute the values

T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

W=\frac{1}{2}m(v^2-v^2_0)

W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

W=0.31 J

4 0
3 years ago
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Rearrange the equation to solve for the permeability of free space (μ0). Remember that the slope is the ratio of magnetic field
IRINA_888 [86]

Complete Question

The complete question is shown on the first uploaded image

Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %

Answer:

The permeability of free space is \mu_0 = 1.32*10^{-6} \ Tm/A

The percentage error is  % error = 5.25%

Explanation:

From the question we are told that

            The slope is  s= 3.9\  G/A , and

Generally  1 \ gauss =  10^{-4 } tesla

               So  s = 3.9 *10^ {-4} T /A

From the relation given in the question

                        \mu_0 = \frac{2R}{N} [\frac{B}{I} ]

 Where R is the radius of the coil  =\frac{Diameter \ of \ coil }{2} = 0.017m

             N is the number of loops of the coil = 10

Now from the question we are told that

                  s = \frac{B}{I}

substituting into the equation above

               \mu_0 = \frac{2R }{N} s

Substituting values

              \mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}

                   = 1.326 *10^ {-6} \ Tm/A

Generally the % error is mathematically represented as

                    %error = \frac{Measured \  value - accepted \ value}{accepted \  value } *100

Given that the accepted value is \mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A

      Hence substituting values

                      %error  = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100

                                  = 5.24

6 0
4 years ago
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