Study the diagram. Point C identifies the ____

Please help I don't know how to answer these questions!

Yuki888 [10]

1) The potential energy is the most at the highest position and the least at the equilibrium position

2) The kinetic energy is the most at the equilibrium position and the least at the highest position

Explanation:

1)

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:

$PE\propto h$

This means that:

The potential energy is the most when the bob is at the highest position
The potential energy is the least when the bob is at the equilibrium position, which is the lowest position

2)

We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,

$E=KE+PE=const.$

where KE is the kinetic energy.

From the equation above, we observe that:

When PE is maximum, KE must be at minimum
When PE is minimum, KE must be maximum

Therefore, this implies that:

The kinetic energy is the most when the potential energy is the least, i.e. at the equilibrium position
The kinetic energy is the least when the potential energy is the most, i.e. at the highest position

Learn more about kinetic and potential energy:

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#LearnwithBrainly

6 0

3 years ago

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

7 0

3 years ago

There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor

VMariaS [17]

Answer:

F = 7.68 10¹¹ N, θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

cos 45 = F₂₄ₓ / F₂₄

sin 45 = F_{24y) / F₂₄

F₂₄ₓ = F₂₄ cos 45

F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

Fₓ = -F₂₁ + F₂₄ₓ

Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis

F_y = - F₂₃ + F_{24y}

F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

F₂₁ = F₂₃

Let's use Coulomb's law

F₂₁ = k q₁ q₂ / r₁₂²

the distance between the two charges is

r = a

F₂₁ = k q² / a²

we calculate F₂₄

F₂₄ = k q₂ q₄ / r₂₄²

the distance is

r² = a² + a²

r² = 2 a²

we substitute

F₂₄ = k q² / 2 a²

we substitute in the components of the forces

Fx = $- k \frac{q^2}{a^2} + k \frac{q^2}{2 a^2} \ cos 45$

Fx = $k \frac{q^2}{a^2}$ ( -1 + ½ cos 45)

F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)

We calculate

F₀ = 9 10⁹ 4.25² / 0.440²

F₀ = 8.40 10¹¹ N

Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

Fₓ = -5.43 10¹¹ N

remember cos 45 = sin 45

F_y = - 5.43 10¹¹ N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

F = -5.43 10¹¹ (i + j) N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

F = $\sqrt{F_x^2 + F_y^2}$

F = 5.43 10¹¹ √2

F = 7.68 10¹¹ N

in angle is

θ = 45º

7 0

3 years ago

A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40º and fly ano

Kobotan [32]

Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.

Total displacement is ac

Total horizontal displacement = 4+4 cos40 =7.06 km

Total vertical displacement = 4 sin40 =2.57 km

Total displacement $= \sqrt{7.06^2+2.57^2}$ = 7.51 km

7 0

3 years ago

The labeled images each represent the wave patterns found in the electromagnetic wave spectrum. which of these images are correc

Phantasy [73]

Energy E of EM radiation is given by the equation E=hf, where h is Planck's constant and f is frequency. It means energy E and frequency f are proportional so as we increase the frequency, energy also increases. Also, the relationship between the wavelength and frequency is c=λ*f where λ is the wavelength and f is frequency and c is the speed of light. This tells us the wavelength and frequency are inversely proportional. So as we increase the frequency the wavelength is getting smaller. So as we go from left to right the frequency increases, energy also increases and the wavelength is decreasing. Or, on the left side we should have low frequency, low radiant energy, and long wavelength. On the right side we should have high frequency, high radiant energy and low wavelength. That is the third graph.

5 0

3 years ago

Study the diagram. Point C identifies the _____ of the wave.