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MissTica
3 years ago
8

Which statement explains whether or not a person should use an experiment of a ball rolling down a hill as evidence that the for

ce of gravity pulls objects straight down?(1 point) They should use it. Any experiment on Earth involving natural motion will provide direct evidence that gravity pulls objects straight down. They should use it. Any experiment on Earth involving natural motion will provide direct evidence that gravity pulls objects straight down. They should not use it. The experiment has several dimensions of motion, including one parallel to the ground, so the motion of the ball does not directly test the force of gravity. They should not use it. The experiment has several dimensions of motion, including one parallel to the ground, so the motion of the ball does not directly test the force of gravity. They should use it. The experiment shows the ball falls toward the bottom of the hill, so it provides direct evidence that gravity pulls an object downward. They should use it. The experiment shows the ball falls toward the bottom of the hill, so it provides direct evidence that gravity pulls an object downward. They should not use it. Gravity should be tested in space, and this experiment could only be tested on Earth.
Physics
1 answer:
Ksenya-84 [330]3 years ago
8 0

The ball raised up the hill is given gravitational potential energy, which

causes it to roll spontaneously down the hill.

The statement that explains whether the ball rolling down the hill

experiment should be used is; <u>The experiment shows the ball falls toward </u>

<u>the bottom of the hill, so it provides direct evidence that gravity pulls an </u>

<u>object downward. They should use it.</u>

Reason:

The equation for the velocity of a ball falling down a hill is v = \sqrt{\dfrac{10}{7} \cdot g \cdot h}

Therefore, by increasing the height from which the ball rolls down the hill,

the velocity of the ball increases, which indicates that the ball is

accelerating, and therefore, being acted on by a force, the force of gravity.

Therefore;

Given that the ball falls or moves with increasing speed towards

the bottom of the hill, which shows that the motion is due to Earth's pull

known as the gravitational force, the experiment should be used.

Learn more here:

brainly.com/question/3175380

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Under the assumption that the beam is a rectangular cantilever beam that is free to vibrate, the theoretical first natural frequ
BartSMP [9]

Answer:

a) Δf = 0.7 n , e)   f = (15.1 ± 0.7) 10³ Hz

Explanation:

This is an error about the uncertainty or error in the calculated quantities.

Let's work all the magnitudes is the SI system

The frequency of oscillation is

        f = n / 2π L² √( E /ρ)

where n is an integer

Let's calculate the magnitude of the oscillation

       f = n / 2π (0.2335)² √ (210 10⁹/7800)

       f = n /0.34257 √ (26.923 10⁶)

       f = n /0.34257    5.1887 10³

       f = 15.1464 10³ n

a) We are asked for the uncertainty of the frequency (Df)

       Δf = | df / dL | ΔL + df /dE ΔE + df /dρ Δρ

in this case no  error is indicated in Young's modulus and density, so we will consider them exact

       ΔE = Δρ = 0

       Δf = df /dL  ΔL

       df = n / 2π   √E /ρ   | -2 / L³ | ΔL

       df = n / 2π 5.1887 10³ | 2 / 0.2335³) 0.005 10⁻³

       df = n 0.649

Absolute deviations must be given with a single significant figure

        Δf = 0.7 n

b, c) The uncertainty with the width and thickness of the canteliver is associated with the density

 

In your expression there is no specific dependency so the uncertainty should be zero

The exact equation for the natural nodes is

          f = n / 2π L² √ (E e /ρA)

where A is the area of ​​the cantilever and its thickness,

In this case, they must perform the derivatives, calculate and approximate a significant figure

        Δf = | df / dL | ΔL + df /de  Δe + df /dA  ΔA

        Δf = 0.7 n + n 2π L² √(E/ρ A) | ½  1/√e | Δe

               + n / 2π L² √(Ee /ρ) | 3/2 1√A23  |

the area is

        A = b h

        A = 24.9  3.3  10⁻⁶

        A = 82.17 10⁻⁶ m²

        DA = dA /db ΔB + dA /dh Δh

        dA = h Δb + b Δh

        dA = 3.3 10⁻³ 0.005 10⁻³ + 24.9 10⁻³ 0.005 10⁻³

        dA = (3.3 + 24.9) 0.005 10⁻⁶

        dA = 1.4 10⁻⁷ m²

let's calculate each term

         A ’= n / 2π L² √a (E/ρ A) | ½ 1 /√ e | Δe

         A ’= n/ 2π L² √ (E /ρ)      | ½ 1 / (√e/√ A) |Δe

        A ’= 15.1464 10³ n ½ 1 / [√ (24.9 10⁻³)/ √ (81.17 10⁻⁶)] 0.005 10⁻³

        A '= 0.0266  n

        A ’= 2.66 10⁻² n

       A ’’ = n / 2π L² √ (E e /ρ) | 3/2  1 /√A³ |

       A ’’ = n / 2π L² √(E /ρ) √ e | 3/2  1 /√ A³ | ΔA

       A ’’ = n 15.1464 10³ 3/2 √ (24.9 10⁻³) /√ (82.17 10⁻⁶) 3 1.4 10⁻⁷

       A ’’ = n 15.1464 1.5 1.5779 / 744.85 1.4 10⁴

       A ’’ = 6,738 10²

we write the equation of uncertainty

     Δf = n (0.649 + 2.66 10⁻² + 6.738 10²)

The uncertainty due to thickness is

    Δf = 3 10⁻² n

The uncertainty regarding the area, note that this magnitude should be measured with much greater precision, specifically the height since the errors of the width are very small

     Δf = 7 10² n

 d)    Δf = 7 10² n

e) the natural frequency n = 1

       f = (15.1 ± 0.7) 10³ Hz

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Which of the following is a simple machine? A. wedge B. meat grinder C. car D. bicycle
kati45 [8]
I think the answer is D. Bicycle
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A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p
alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

F=qvB

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

qvB = \frac{mv^2}{r}

which can be rewritten as

v=\frac{qB}{mr}

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

\frac{2\pi r}{T}=\frac{qB}{mr}

So, we get:

T=\frac{2\pi m r^2}{qB}

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) a. f/10

The frequency of revolution of a particle in uniform circular motion is

f=\frac{1}{T}

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}

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_________ is a layer of the earth that is classified not by composition
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Answer:

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The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.

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The first element on the periodic table of elements is helium true of false
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Answer:

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