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Dennis_Churaev [7]

3 years ago

6

How do observing and inferring differ?

2 answers:

ddd [48]3 years ago

8 0

Observation = you are seeing it as in: I can physically see rain therefore it IS raining
Inference = making a point based on suggestions as in: there are puddles on the ground. It must be raining therefore it IS raining

Illusion [34]3 years ago

6 0

Observing is commenting on what you see e.g: It is raining.
Inferring is drawing conclusions from observations for example you see the road is wet so you might infer that it rained last night.

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The current through a 10 ohm resistor connected to a 120 volt power supply is

Leno4ka [110]

Answer:I=12 A

Explanation:

Given

Resistance $R=10 \Omega$

Voltage $V=120 V$

According to ohm's law current through a conductor is directly proportional to the voltage applied.

$V\propto I$

$V=IR$

where V=Voltage

I=Current

R=resistance

$I=\frac{V}{R}$

$I=\frac{120}{10}$

$I=12 A$

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What type of electromagnetic wave does a light bulb and a radio antenna release?

andrew-mc [135]

I believe that a light bulb releases visible light and a radio antenna releases a radio waves

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sergij07 [2.7K]

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3 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec

nadezda [96]

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

$Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\$

On substituting the above values we get the required acceleration

$Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}$

Therefore,the acceleration of the train is 5 m/s².

4 0

3 years ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

Over [174]

Answer:

$4.1\cdot 10^8 N$

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

$g=9.8 m/s^2$ is the gravitational acceleration

$h=11,000 m$ is the depth

Substituting,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

And so, the inward force exerted on it is

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

3 years ago

Read 2 more answers