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3 years ago

16. How much force would it take to accelerate a 75kg object by 5 m/szą

Physics

1 answer:

patriot [66]3 years ago

7 0

Answer:

The force experienced is 375 N.

Explanation:

Here,

Mass of the object( M)= 75 kg

Acceleration of the body(a) =5 m/s²

Force required to accelerate the body( F) =?

Then, we know ,

From newton's second law of motion,

$F=M*a$

= $75* 5$

=375 kgm/s²

= 375 Newton

So, The force experienced by the 75 kg body that undergoes an acceleration of 5 m/s² is 375 N.

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4 years ago

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Read 2 more answers

Si un movil parte del reposo logrando una aceleracion de 5 metros por segundo al cuadrado durante 8 segundos calcular la velocid

Komok [63]

Answer:

Velocidad final, V = 40 m/s

Explanation:

Dados los siguientes datos;

Aceleración = 5 m/s²

Velocidad inicial = 0 m/s (ya que comienza desde el reposo)

Tiempo = 8 segundos

Para encontrar la velocidad final, usaríamos la primera ecuación de movimiento;

$V = U + at$

Dónde;

V es la velocidad final.
U es la velocidad inicial.
a es la aceleración.
t es el tiempo medido en segundos.

Sustituyendo en la fórmula, tenemos;

$V = 0 + 5*8$

$V = 0 + 40$

$V = 40$

<em>Velocidad final, V = 40 m/s</em>

6 0

3 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/sm/s . When she plays the note A, it is in perfect tun

dezoksy [38]

Answer:

A) beats per second she will hear if she now plays the note A as the tuning fork is sounded = 5.13 beats per second

B) length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork = 0.0045m

Explanation:

A) First of all, wavelength = v/f

Where v is speed of wave and f is frequency.

Thus, wavelength of the sound wave of Note A is;

f2 = 440 Hz and v = 342m/s

λ = 342/440 = 0.7773m

Now, since the air inside the note was warmed after a while, the wave will will have a new frequency which we'll call (f1) and and new speed (v'), thus;

f2 = v'/λ = 346/0.7773 = 445.13 Hz

Now let's calculate beat frequency(fbeat).

fbeat = (f1 - f2)

So fbeat = 445.13 - 440 = 5.13Hz or 5.13 beats per second

B) Now, frequency of standing wave models (fm) = n(v/2L)

Where n is a positive integer and L is the open tube length

Making L the subject of the formula, we have; L = nv/2fm

Now from earlier derivation, we see that v = fλ and in this case, v=fλ

Thus, let's replace v with fλ to het;

L = nλ/2

If we take, n=1, L = (1 x 0.7773)/2 = 0.3887m

Now, when the air inside the tube has warmed, it will have a new length to eliminate beats and give same frequency of 440Hz.

So let's call this new length L1;

So L1 = v'/2(f2) = 346/(2x440) = 346/880 = 0.3932m

So the length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork will be;

ΔL = L1 - L = 0.3932 - 0.3887 = 0.0045m

5 0

4 years ago

The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center. If

zvonat [6]

Answer:

The distance between their eye and the ceiling is 33.06 m.

Explanation:

Given that,

The ceiling of a large symphony hall is covered with acoustic tiles which have small holes that are 4.35 mm center to center, D = 4.35 mm

Diameter of the eye of pupil, d = 5.1 mm

Wavelength, $\lambda=550\ nm$

We need to find the distance between their eye and the ceiling. Using Rayleigh criteria, we get the distance as follows :

$L=\dfrac{Dd}{1.22\lambda}\\\\L=\dfrac{4.35\times 10^{-3}\times 5.1\times 10^{-3}}{1.22\times 550\times 10^{-9}}\\\\L=33.06\ m$

So, the distance between their eye and the ceiling is 33.06 m.

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3 years ago