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Dovator [93]
2 years ago
12

A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?

Physics
1 answer:
Sveta_85 [38]2 years ago
5 0
A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N
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A helium-filled balloon occupies a volume of 15 cubic meters at sea level. the balloon is released and raises to a point in the
Elena L [17]

According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.

Therefore,

P_{1} V_{1} =P_{2} V_{2}

Given P_{1} = 1 atm , V_{1} = 15 \ cubic \ meter and P_{2} = 0.75\ atm.

Thus,

V_{2} = \frac{P_{1}\times V_{1}  }{ P_{2} } = \frac{1\times 15}{0.75} =20 m^3

5 0
3 years ago
Un engrane que gira con una velocidad de 20 rad/s, es acelerado durante 5 segundos hasta alcanzar una velocidad de 35 rad/s
larisa [96]

Answer:

a) La aceleración angular es: \alpha=2\: rad/s^{2}

b) El engranaje gira 125 radianes.

c) El engranaje hara aproximadamente 20 revoluciones.

Explanation:

a)

La aceleración angular se define como:

\alpha=\frac{\Delta \omega}{\Delta t}

Donde:

  • Δω es la diferencia de velocidad angular (en otras palabras ω(final)-ω(inicial))
  • Δt es el tiempo en el que occure el cambio de velocidad angular

\alpha=\frac{35-25}{5}

\alpha=2\: rad/s^{2}

b)

El desplazamiento angular puede ser calculado usando la siguiente ecuación:

\theta=\theta_{i}+\omega_{i}t+\frac{1}{2}\alpha t^{2}

Aqui el angulo inicial es 0, por lo tanto.

\theta=20(5)+\frac{1}{2}(2)(5)^{2}

\theta=125\: rad

El engranaje gira 125 radianes.

c)

Lo que debemos hacer aquí es convertir radianes a revoluciones.

Recordemos que 2π rad = 1 rev

Entonces:

\theta=125\: rad \times \frac{1\: rev}{2\pi\: rad}=19.89\: rev

Por lo tanto el engranaje hara aproximadamente 20 revoluciones.

Espero te haya sido de ayuda!

6 0
3 years ago
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

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Period is the answer

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