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4 years ago

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Identify the examples of values. Check all that apply. Inez thinks it is important to be honest with her friends. Molly was rais

ed in a foreign country. Byron tries to treat all people with respect. Lydia cares about expressing her creativity. Armando enjoys playing basketball. Marc would like to learn more about fashion.

2 answers:

Lubov Fominskaja [6]4 years ago

5 0

Answer:

Inez thinks it is important to be honest with her friends.

Byron tries to treat all people with respect.

Lydia cares about expressing her creativity.

Explanation:

Just did on Edge

valentinak56 [21]4 years ago

3 0

Answer:

1 3 4

Explanation:

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A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0

4 years ago

Read 2 more answers

Pls help !!!!!!!!!!!!!!!!!!!!!!!!!!!!

Anna11 [10]

Answer:

B. Identifying the types of data to be gathered

Explanation:

Identifying the types of data to be gathered is one of the part of designing a set of experimental procedure.

An experimental procedure is a methodical process that must be followed in order to attain the goal of the experiment.

As with most experiments, data is collected for further analysis and interpretation.
The choice of way of collecting data and the type of data to be collected is very important when designing an experiment.

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3 years ago

Describe the movement of air over 2 nearby land areas, one of which is heated more than the other.

77julia77 [94]

In this case, the air from the warm area will always start moving towards the colder areas because as the temperature in both lands should be equal. This is one of the laws of thermodynamics.

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4 years ago

A girl holds an arrow in her hands, ready to shoot the arrow at a dragon who is just outside her arrow’s range of 45 meters. The

gregori [183]

First, assume that the arrow leaves the bow with a velocity of 45 m/s above the horizontal with respect to the bow.

Since the bow is moving at 8.5 m/s 45º below the horizontal, find the initial velocity of the arrow with respect to the ground:

$\vec{v}_{AG}=\vec{v}_{AB}+\vec{v}_{BG}$

This equation reads:

<em>The velocity of the arrow with respect to the ground </em><em>is equal to</em><em> the velocity of the arrow with respect to the bow </em><em>plus</em><em> the velocity of the bow with respect to the gound.</em>

Notice that this is a vector equation. Then, the vertical and horizontal components of the velocities must be added separately:

$\begin{gathered} v_{AG-x}=v_{AB-x}+v_{BG-x} \\ v_{AG-y}=v_{AB-y}+v_{BG-y} \end{gathered}$

Find the vertical and horizontal components of the velocity of the arrow with respect to the bow and the velocity of the bow with respect to the ground:

$\begin{gathered} v_{AB-x}=v_{AB}\cos (\theta) \\ =45\frac{m}{s}\cdot\cos (30º) \\ =38.97\frac{m}{s} \end{gathered}$

$\begin{gathered} v_{AB-y}=v_{AB}\sin (\theta) \\ =45\frac{m}{s}\sin (30º) \\ =22.5\frac{m}{s} \end{gathered}$

Similarly, for the velocity of the bow with respect to the ground:

$\begin{gathered} v_{BG-x}=6.01\frac{m}{s} \\ v_{BG-y}=-6.01\frac{m}{s} \end{gathered}$

Then, the vertical and horizontal components of the initial velocity of the arrow with respect to the ground, are:

$\begin{gathered} v_{AG-x}=38.97\frac{m}{s}+6.01\frac{m}{s}=44.98\frac{m}{s} \\ \\ v_{AG-y}=22.5\frac{m}{s}-6.01\frac{m}{s}=16.49\frac{m}{s} \end{gathered}$

Use the horizontal component of the velocity to find how long it takes for the arrow to travel a horizontal distance x of 55 meters. Then, use that time to find the vertical position of the arrow.

Since the horizontal movement of the arrow is uniform, then:

$v_{AG-x}=\frac{x}{t}_{}$

Isolate t and substitute x=55m, v_{AG-x}=44.98 m/s:

$\begin{gathered} t=\frac{x}{v_{AG-x}} \\ =\frac{55m}{44.98\frac{m}{s}} \\ =1.2227s \end{gathered}$

The vertical motion of the arrow is a uniformly accelerated motion. Then, the vertical position is given by:

$y=v_{AG-y}t-\frac{1}{2}gt^2$

Replace v_{AG-y}=16.49 m/s, t=1.2227s and g=9.81 m/s^2 to find the vertical position of the arrow when the horizontal position is 55 meters. This matches the elevation of the dragon with respect to the girl when the girl shoots:

$\begin{gathered} y=(16.49\frac{m}{s})(1.2227s)-\frac{1}{2}(9.81\frac{m}{s^2})(1.2227s)^2 \\ =12.829\ldots m \\ \approx12.8m \end{gathered}$

Therefore, the dragon is 12.8 meters above the girl when the arrow is shoot.

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1 year ago

Our two intrepid relacar drivers are named Pam and Ned. We use these names to make it easy to remember: measurements made by Pam

kenny6666 [7]

This is a problem based on the logic and interpretation of the variables. From the measured data taken

what is collected by the two individuals is expressed as,

- NED reference system: (x, t)

- PAM reference system: (x ', t')

From the reference system we know that ν is the speed of PAM (the other reference system) as a measurement by NED.

Then ν' is the speed of NED (from the other system of the reference) as a measurement by PAM.

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3 years ago