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MAVERICK [17]
3 years ago
8

Rheumatoid arthritis is a disorder in which the joints swell and cause pain. Rheumatoid arthritis causes the body’s own immune c

ells to attack the membrane of the joints. Which body system does rheumatoid arthritis affect?
nervous



musculoskeletal



integumentary



excretory



circulatory
Physics
2 answers:
Dafna11 [192]3 years ago
7 0
This would be the musculoskeletal 
olga55 [171]3 years ago
3 0

It would be the musculoskeletal because the disorder focuses on the joints which are a part of the musculoskeletal body system.

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A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
statuscvo [17]

Answer:

37357 sec  

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J  

Energy burns in 1 lift = m g h

                                  = 80 x 9.8 x 1 = 784 J

lifts required = \frac{(14.644 x 10^6)}{784}

                      = 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec  

or 622 min

or 10.4 hrs

3 0
3 years ago
The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di
nlexa [21]

Answer and Explanation:

Let:

x(t)=Acos(\omega t+ \phi)

The equation representing a simple harmonic motion, where:

x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

\phi = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

8 0
3 years ago
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A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter st
devlian [24]

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

4 0
3 years ago
A car travels 30 km north in 25 min. and 40 km east in 35 min. What is the total time in hours? Be careful to carry over the pro
inna [77]
25+35=60mins so 1.0 hours
3 0
3 years ago
Read 2 more answers
Listed below are the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones. Use the given data to const
Fantom [35]

Answer:

The 5-number summary is

1. Median = 0.93 W/kg

2. Lower quartile = 0.69 W/kg

3. Upper quartile = 1.16 W/kg

4. Minimum value = 0.54 W/kg

5. Maximum value = 1.42 W/kg

Explanation:

We are given the measured radiation absorption rates​ (in W/kg) corresponding to 11 cell phones.

1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57

What is 5-number summary?

A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.

These statistical characteristics are:    

1. Median

2. Lower quartile

3. Upper quartile  

4. Minimum value  

5. Maximum value  

1. Median:

Arrange the data in ascending order

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

(n+1)/2 gives the median value of the data set.

(11 + 1)/2 = 6th position

Therefore, 0.93 W/kg is the median of the data set.

2. Lower quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The lower quartile is the median of the lower half of the data set.

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

The median is 6/2 = 3rd position

Therefore, the lower quartile of the data set is 0.69 W/kg

3. Upper quartile:

Divide the data set into two equal halfs (include median in both if n = odd)

Lower half = 0.54 0.57 0.69 0.75 0.85 0.93

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The upper quartile is the median of the lower half of the data set.

Upper half = 0.93 0.95 1.16 1.17 1.18 1.42

The median is 6/2 = 3rd position

Therefore, the upper quartile of the data set is 1.16 W/kg

4. Minimum value:

The minimum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the minimum value of the data set is 0.54 W/kg

5. Maximum value  

The maximum value is the least value in the data set.

0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42

Therefore, the maximum value of the data set is 1.42 W/kg

The box plot is illustrated in the attached diagram.

6 0
3 years ago
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