A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?

Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198

Answers

Answer 1

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

Answer 2

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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the second option - B

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Romeo is a 68 kg astronaut. Juliet is a beautiful cosmonaut who is standing on the balcony of a 4.58 x 10^5kg space station that is at rest and out of gas. Romeo is floating 25 meters away from the space station’s center of mass, how strong is the force between Romeo and Juliet?

Answers

Answer:

F =  3.32 x 10⁻⁶ N

Explanation:

The force of attraction between two masses is given by Newton's Law of Gravitation, as follows:

F = Gm₁m₂/r²

where,

F = Force between Romeo and Juliet = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Romeo = 68 kg

m₂ = mass of space station = 4.58 x 10⁵ kg

r = distance = 25 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(68 kg)(4.58 x 10⁵ kg)/(25 m)²

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Please help me with this question guys.

Answers

Answer:

The average speed is 22.2 km/h

Explanation:

Average Speed

Given an object travels a total distance d and took a total time t, then the average speed is:

[tex]\displaystyle \bar v=\frac{d}{t}[/tex]

The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:

[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]

Then he drives d2=7 km at v2=43 km/h taking a time of:

[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]

The total time is

t=0.467 h + 0.163 h = 0.63 h

The total distance is

d = 7 km + 7 km = 14 km

The average speed is:

[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]

The average speed is 22.2 km/h

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Answers

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Our school needs to offer healthier options in the lunchroom. Elever High School has recently updated its cafeteria menu to include whole wheat pasta and breads, a fresh salad bar, and other healthy menu items. Students there claim that they have more energy and focus throughout their school day. Let's encourage healthier menus in our lunchroom!

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Answers

Answer:

357 J

Explanation:

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The kinetic energy of the arrow after it has been shot is given by half of the product of the arrow's mass and velocity of the arrow.

Here there are no other forms of energy at play here. Only potential and kinetic energy.

As we know that in any system the energy is conserved accordingly the elastic potential energy of the arrow will be equal to the kinetic energy of the bow after it is released i.e., 357 J.

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Answers

It is the proportion predicted to be present in the early universe.

The hydrogen and helium abundance helps us to model the expansion rate of the early universe.

In the abundance of hydrogen and helium, we can say that they account for nearly all the nuclear matter in today's universe.

In big Bang model, the universe is mostly light or protons.

This abundance of hydrogen and helium is consistent with this big bang model. The process of forming this hydrogen and helium is often called big bang nucleosynthesis.

The Schramm's model for relative abundances indicate that helium is about 25% by mass and hydrogen about 73% with all other elements constituting less than 2%.

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