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2 years ago

The sun rose at 6:54 this morning.is this an observation or inference?

Physics

2 answers:

Levart [38]2 years ago

8 0

Answer:

observation

Explanation:

Through an observation information is acquired. The acquiring of information must be through the five senses directly or indirectly. When a person takes an observation using eyes it is a form direct observation. When a person takes an observation using scientific instruments it is a form indirect observation.

Inference is the deduction of observations taken.

Here, the sun rose at 6:54 this morning is an observation i.e., a person saw the sun rise at 6:54. It has not been deduced that the sun rose at 6:54.

jeka57 [31]2 years ago

4 0

Observation: You saw/ observed something and you are stating that.

Inference: You are predicting something to be true. If so and so, then this will occur.

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: In heavy rushIn heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5

zhuklara [117]

Guessing you want the average speed. We can multiple each speed by the time we spent going that speed, and them all together and then divide by the total time we spent in traffic to get the average speed. We spent a total of 7.5 minutes in traffic, so average speed = (12*1.5+0*3.5+15*2.5)/7.5 = 7.4 m/s

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2 years ago

the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s

12345 [234]

The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2 π + 3cos 2 π - ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

4 0

1 year ago

What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?

ratelena [41]

= 80 N/40 kg
= 2 m/s 2

4 0

2 years ago

A 9,100-microfarad [muF] capacitor is charged to 22 volts [V]. If the capacitor is completely discharged through an iron rod

MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

2 years ago