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faltersainse [42]

3 years ago

9

When you weigh yourself on good old terra firma (solid ground), your weight is 540 N. In an elevator your apparent weight is 485

N. What is the magnitude of the elevator’s acceleration?

1 answer:

hichkok12 [17]3 years ago

8 0

Type it in safari and see

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Why is it important for professionals in any field to be accurate and precise with their data collection??

kobusy [5.1K]

Answer:When prfessionals take data collections its important becasue it can cause error. Lets say they are sloppy with thier work and end up getting something that is not near what should be happening. This can have a major affect on the truth of what they are doing and an effect on thier end result in general.

Explanation:

6 0

3 years ago

Read 2 more answers

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

Power selection feature for resistors to become water modules 10 liters of water at 25°C to đến

Mila [183]

Answer:

P = 2439.5 W = 2.439 KW

Explanation:

First, we will find the mass of the water:

Mass = (Density)(Volume)

Mass = m = (1 kg/L)(10 L)

m = 10 kg

Now, we will find the energy required to heat the water between given temperature limits:

E = mCΔT

where,

E = energy = ?

C = specific heat capacity of water = 4182 J/kg.°C

ΔT = change in temperature = 95°C - 25°C = 70°C

Therefore,

E = (10 kg)(4182 J/kg.°C)(70°C)

E = 2.927 x 10⁶ J

Now, the power required will be:

$Power = P = \frac{E}{t}$

where,

t = time = (20 min)(60 s/1 min) = 1200 s

Therefore,

$P = \frac{2.927\ x\ 10^6\ J}{1200\ s}$

<u>P = 2439.5 W = 2.439 KW</u>

7 0

2 years ago

Whenever Dylan walks into his room, his roommate loudly blows an air horn, which startles Dylan. Dylan doesn't think this is fun

Sav [38]

Answer:

stimuli

Explanation:

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3 years ago

I need the answer for both of the questions please

Lady_Fox [76]

But even more pain on pain and then pain and pain ya feel me and even more pain okay and yes more pain

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3 years ago