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max2010maxim [7]

3 years ago

11

3. Consider a locomotive and the rest of a freight train to be a single object. Suppose the locomotive is pulling the train up a

hill. Describe the action and reaction forces that cause the locomotive to move up the hill, such as the reaction force and gravity.
This question has been posted before put the answers were not correct.

1 answer:

Nonamiya [84]3 years ago

6 0

Answer:

Action - Pulling up the train.

Reaction - Friction on the locomotive

Explanation:

Locomotive is pulling the train upwards ,

Which is the action force applied by the locomotive,

As a reaction locomotive will be pulled by the train which is the reaction of pulling

Now, considering it as a action on locomotive , friction force will act on it as a reaction upwards which will result to move it upwards.

For train action is pulling up by locomotive and reaction will be friction acting on it downwards.

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The universe was 5 percent its current size when light left objects observed now at redshift of ______________

Nuetrik [128]

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.

Reasons:

The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

$\displaystyle \frac{a}{a_0} =\mathbf{ \frac{1}{1 + z}}$

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

$\displaystyle \frac{a}{a_0} =5\% = 0.05= \frac{1}{1 + z}$

$\displaystyle 0.05 = \frac{1}{1 + z}$

0.05 + 0.05·z = 1

$\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19$

The redshift is of the observed light is, z = <u>19</u>

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4 0

2 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

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3 years ago

What is the correct equation for calculating the average atomic mass for 3 isotopes? (pls be 100%of your answer pls no guessing)

mariarad [96]

<u>Answer:</u>

<em>The correct equation for measuring the average microscopic weight for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>

<u>Explanation:</u>

To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.

Take the correct weight of each isotope (that will be in decimal form)

Multiply the weight of each isotope by its abundance

Add each of the results together.

<em>This gives the required average microscopic weight of the three isotopes.</em>

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Answer:

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