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4 years ago

Which factors could be potential sources of error in the experiment? check all that apply.

Physics

2 answers:

Vadim26 [7]4 years ago

6 0

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

WARRIOR [948]4 years ago

6 0

Answer:

The Correct Answer is A, C, and E.

Explanation:

There are many potential sources of errors when designing and carrying pout experiments.

Personal error: Inaccurate observation by observers, If the data collecting has two observers and they have collected two different data which is not matching with the experiment data then it would be considered as the Observer or personal error.
Instrumental errors: If Calibration of the instruments is not carried out correctly then it will produce incorrect data then it will be categorized under Instrumental error of experiment.

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If you are doing a "rowing" motion, what muscle of the shoulder are you using?

Andrej [43]

The upper back muscles being worked while using a rowing machine .your upper trapezius and rhomboids located between your shoulder blades, and latissimus dorsi located beneath the armpits

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3 years ago

If you were thinking about a washing machine as a system which of the following represents the inputs?

Lostsunrise [7]

Answer:

The answer is a, the dirty cloths, water and detergent.

Explanation:

The answer is the above selected because the inputs basically represent the data that are passed through the system to generate the output.

In this case, the inputs are the aforementioned in the answer while the possible output would literally be the clean cloths.

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3 years ago

Which type of mirror would you use to cook a hot dog and why?

alexandr1967 [171]

As long as it’s a good mirror then any one of them is fine bc at the end of the day i’m getting a hot dog

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3 years ago

) By observing that the centripetal acceleration of the Moon around the Earth is ac = 2.7 × 10-3 m/s2, what is the gravitatonal

Sedbober [7]

Answer:

G = 6,786 10⁻¹¹ m³ / s² kg

Explanation:

The law of universal gravitation is

F = G m M/ r²

Where G is the gravitational constant, m and M are the masses of the bodies and r is the distance from their centers

Let's use Newton's second law

F = m a

The acceleration is centripetal

a = $a_{c}$

We replace

G m M / r² = m $a_{c}$

G = $a_{c}$ r² / M

Let's replace and calculate

G = 2.7 10⁻³ (3.88 10⁸)² / 5.99 10²⁴

G = 6,786 10⁻¹¹ m³ / s² kg

Let's perform a dimensional analysis

[N m²/kg²] = [kg m/s² m² / kg²] = [m³ / s² kg]

4 0

3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

3 years ago