Answer:
a. 60.5 kg
Explanation:
Given data,
The maximum water a boat can displace is, 60.5 ml
According to the principle of buoyancy, the weight of the floating body is equal to the weight of the liquid displaced.
Under standard temperature and pressure, a unit mass of water equals one liter.
If a boat can displace a maximum of 60.5 ml of water, then it can hold a mass of a maximum of 60.5 kg of mass.
Answer:
La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.
Explanation:
El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.
La distancia recorrida, x
, por un móvil que tiene un MRU con un velocidad v durante el intervalo de tiempo t es:
x= x0 + v*t
donde x0 es la posición inicial.
En este caso:
Reemplazando:
x= 22 m + 5 m/s* 30 s
Resolviendo:
x= 22 m + 150 m
x= 172 m
<u><em>La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.</em></u>
The loudness of the sound at the rock concert, where the intensity of the sound is1 x 10⁻¹ Wm⁻² is 110 dB.
Here we are dealing with loudness which is the perception of the Intensity of the sound.
The formula to refer to in order to find the value of the loudness of a sound is ,
db= 10log(I/I₀)
As we are provided with the current intensity which is 1 x 10⁻¹ Wm⁻². and the initial intensity which is 1 x 10⁻¹² Wm⁻².
So, by substituting the required values in the formula we get
db= 10 * log( 1 x 10⁻¹ /1 x 10⁻¹²)
= 10 * 11 log(10)
= 110
So, the result is 110 dB.
To know more about the intensity of sound refer to the link brainly.com/question/9323731?referrer=searchResults.
To know more about questions related to loudness refer to the link brainly.com/question/21094511?referrer=searchResults.
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To find the force we use the formula,
F = ma , where m is mass and a acceleration
Using the formula,
F = ma
F = 0.42 x 14.8
F = 6.216 N / 6.22 N
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(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
<h3>
Work done in the spring</h3>
The work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
W(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
W(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
Learn more about work done here: brainly.com/question/25573309
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