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4 years ago

13

Parallax error occurs when the observer records data when he/she is at an angle to the event he/she is observing. Where do you t

hink the observer should be to avoid the parallax error?

1 answer:

kompoz [17]4 years ago

4 0

Answer:Orient your line of sight directly above the measurement markings.

Explanation: parallax error is a type of systematic error that occurs when an observer views a measurement marking at a wrong angle. This causes a noticeable disparity in results obtained. Therefore the best way to prevent this error is to view and record the data from the correct angle. This can be obtained by:

-Place the measurement device on its edge so it is level with the object being measured.

-Seek out the finest possible edge of the measurement device, or use a device with finer edges.

In conclusion, ask other observers to also take the reading and get an average of their results. It can help cancel out parallax error results.

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adoni [48]

450x force to stop her 5 seconds

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4 years ago

Fine grains of beach sand are assumed to be spheres of radius 64.8 µm. These grains are made of silicon dioxide which has a dens

Nutka1998 [239]

Answer:

1) Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) 0.08901 kg of sand would have surface area equal the surface area of the cube.

Explanation:

1) Radius of the grain,r = 64.8 µm = $6.48\times 10^{-5}m$

Volume of the sphere = $\frac{4}{3}\pi r^3$

Volume of the grain of a sand:

$V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3$

$V=1.1397\times 10^{-12} m^3$

Density of a grain of sand = $d=2600 kg/m^3$

Mass of a grain of a sand = M

$d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}$

$M=2.9632\times 10^{-9} kg$

Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) Surface are of sphere: $4\pi r^2$

Surface area of a grain:

$A=4\times 3.14\times (6.48\times 10^{-5}m)^2$

$A=5.2766\times 10^{-8} m^2$

Length of the cube = a = 0.514 m

Total surface area of cube ,A'= $6a^2$

$A'=6\times (0.514 m)^2=1.5851 m^2$

let the number grains with area equal to total surface area of cube be x.

$A'=A\times x$

$x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7$

Volume of x number of grains :V'

$V'=V\times x$

$V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7$

$V'=3.4236\times 10^{-5} m^3$

Mass of $3.4236\times 10^{-5} m^3$ of sandL:

= $3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg$

0.08901 kg of sand would have surface area equal the surface area of the cube.

8 0

3 years ago

Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad

Anettt [7]

Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, $d=1.9\times 10^7\ m$

Mass of Earth, $M=5.97\times 10^{24}\ kg$

Radius of Earth, $r=6.37\times 10^6\ m$

We need to find the force of gravity above the surface of Earth. It is given by :

$F=G\dfrac{mM}{R^2}$

R = r + d

R = 25370000 m

$F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}$

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

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