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charle [14.2K]
3 years ago
5

The temperatures (in degrees Fahrenheit) in Long Island recorded by the weather bureau over a week were 42, 49, 53, 55, 50, 47,

and 52. Which measure should the weather bureau use to calculate how far apart the upper and lower quartiles of the week's daily temperatures were?
Physics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

Inter Quartile Range

Explanation:

Quartile is a positional statistical average, which divided the data into 4 equal halves.

Q1 (Lower Quartile) has 25% data below it, 75% above it. Q3 (Upper Quartile) has 75% data below it, 25% above it.

Interquartile range is the measure used to calculate how far the lower & upper quartiles are.

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Your mass in kilograms if you weight 170 pounds
Aleonysh [2.5K]
1 pound ≈ 0.4536 kg

170 pounds ≈ 170 * 0.4536 kg

                     ≈ 77.112 kg

                      
5 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
I need help with number 6 and 7 please and I’ll mark brainliest
arsen [322]

Answer:

n.6 is T

Explanation:

because mass always stays the same where ever you are but weight changes depending on the gravity

3 0
3 years ago
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Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their
geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\

The magnitude of the final velocity of the wreck can be found as:

v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0
3 years ago
The image below shows two opposite forces acting on a rope, what can we say is true about the affect of the forces on the rope?
Tamiku [17]
It would be D) the rope is pulled to the right. This is because their is a greater force in that direction.
8 0
3 years ago
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