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3 years ago

What are some potential sources of error in the experiment that is described in the article?

Physics

1 answer:

mezya [45]3 years ago

4 0

Explanation:

common source of error include instrumental, environmental, procedural , and human Police Stop all of these errors can be either random for systematic depending on how they affect the results.

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A single-turn circular loop of wire of radius 65 mm lies in a plane perpendicular to a spatially uniform magnetic field. During

BabaBlast [244]

Answer:

Explanation:

change in flux = no of turns x area of loop x change in magnetic field

= 1 x π 65² x 10⁻⁶ x ( 650 - 350 ) x 10⁻³

= 3.9 x 10⁻³ weber .

rate of change of flux = change of flux / time

= 3.9 x 10⁻³ / .10

= 39 x 10⁻³ V

= 39 mV .

Since the magnetic flux is directed outside page and it is increasing , induced current will be clockwise so that magnetic field is produced in opposite direction to reduce it , as per Lenz's law.

5 0

3 years ago

For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

3 0

4 years ago

A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared to the 10 kg ball, the 1 kg ball has

Zepler [3.9K]

The answer is the second option, or 1/10 the same momentum.

7 0

3 years ago

Read 2 more answers

The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using

Crazy boy [7]

Answer:

6.04788 in

$115.82654\ in^3$

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

$c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in$

Diameter is given by

$d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in$

The diameter is 6.04788 in

$6.04788\times 2.54=15.3616152\ cm$

Volume of sphere is given by

$v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3$

The volume is $115.82654\ in^3$

$115.82654\times \dfrac{1}{1728}=0.06702\ ft^3$

Fall velocity is given by

$V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s$

The velocity of the fall will be 78.38779 m/s

Mass is given by

$m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb$

$1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg$

The mass is 0.88159 kg

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J$

The kinetic energy is 34.55294 J

4 0

3 years ago

A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent

solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

$I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}$

The angular acceleration is given as

$a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}$

The angular velocity is given as

$w=at\\w=(0.905)(2.55)\\w=2.31rad/s$

So the Kinetic Energy is given as

$K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J$

3 0

3 years ago