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3 years ago

What must be part of a quantitative observation?

Physics

2 answers:

inna [77]3 years ago

8 0

a number hope this helps

Pani-rosa [81]3 years ago

5 0

Answer with Explanation:

There are two types of Observation:

1.→ Qualitative Observation : In which things are Selected on the basis of Quality ,like Color,Shape , Size ,Behavior,Nature of a person, etc...

2.→Quantitative Observation: Representation is done with the help of numbers ,Alphabets or Special Characters or by Combining all of them.

Option A: <u>A Change of State:</u> ------It denotes Quality.

Option B:<u> A change in Size :</u> ----It can be both Quality as well as Quantity.For Example Animal loosing and gaining weight can be represented in terms of both numbers as well Quality.For example :50-30,Fat-Thin.

Option C: <u>A number-</u>----It denotes Quantity.

Option D: <u>A Color</u>----It denotes Quality.

⇒A Number is : is must a quantitative observation.

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2 years ago

What is true of transverse waves?

topjm [15]

<h2>Answer:D</h2>

Explanation:

Option A:

Surface waves are neither transverse nor longitudinal.They traverse perpendicularly or parallel to the wave's motion along the interface between different media.

Option B:

Transverse waves vibrate perpendicularly to the direction of the propagation of the wave.

Option C:

Sound is a longitudinal wave.Not a transverse wave.

Option D:

Transverse waves don't require a medium for propagation.But they propagate in medium too.

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2 years ago

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You are driving your motorcycle in a circle of radius 75 m on wet pavement. what is the fastest you can go before you lose tract

stira [4]

On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s

Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.

Maximum velocity of a road with friction is given by the formula,

v = μRg

where, v is the maximum velocity

μ is the coefficient of static friction

R is the radius of the circle road

g is the acceleration due to gravity

Given,

μ = 0.20

R = 75m

g = 9.8m/s²

On substituting the given values in the above formula,

v = 0.20× 75 ×9.8

v = 147m/s

So, the Maximum velocity of the wet road is 147m/s.

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1 year ago

Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po

zvonat [6]

Answer:

$0.293I_0$

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

$I_1=\frac{I_0}{2}$

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

$I_2=I_1 cos^2 \theta$

where

$\theta$ is the angle between the axes of the two polarizer

Here we have

$\theta=40^{\circ}$

So the intensity after the 2nd polarizer is

$I_2=I_1 (cos 40^{\circ})^2=0.587I_1$

And substituting the expression for I1, we find:

$I_2=0.587 (\frac{I_0}{2})=0.293I_0$

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2 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

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3 years ago

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