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Rudik [331]
3 years ago
15

A machine part rotates at an angular speed of 0.76 rad/s; its speed is then increased to 2.96 rad/s using an angular acceleratio

n of 1 rad/s 2 . Find the angle through which the part rotates before reaching this final speed. Answer in units of rad.
Physics
1 answer:
son4ous [18]3 years ago
5 0

Answer:

θ = 4.092 °

Explanation:

Given that

Initial Angular speed ,ωi = 0.76 rad/s

Final angular speed ,ωf= 2.96 rad/s

The angular acceleration ,α = 1 rad/s²

We know that

ωf² = ωi ²+ 2 α θ

θ=Angle rotates

2.96² = 0.76²+ 2 x 1 x θ

\theta=\dfrac{2.96^2-0.76^2}{2}\ degrees\\\theta=4.092\ degrees\\

Therefore the angle rotates by machine part will be 4.092 degrees.

θ = 4.092 °

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Nitially the flame produces radiation<span> which heats the tin can. The tin can then</span>transfers heat<span> to the water </span>through<span> conduction. The hot water then rises to the top, in the convection process. </span>
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Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

mass of vehicle m = 1000 kg

for a low speed test; V = 2.5 m/s

bumper maximum deflection = 4 cm = 0.04 m

First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

hence;

W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

k = 3906250 N/m

Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

3 0
2 years ago
A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?
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F = 400 Pa x 2.5 m2
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2 years ago
Determina el tamaño de la arista de un cubo hecho de plata, cuya masa es de 10.49 kg.
Sidana [21]

Answer:

88.2 N

Explanation:

Datos

Lcubo = 10 cm = 0.1 m

Vcubo = Vfluido desalojado= 0.1 m x 0.1 m x 0.1 m = 10-3 m

mcubo = 10 kg

dfluido = 1000 kg/m3

g = 9.8 m/s2  

Sabemos que el peso aparente de un cuerpo que se sumerge en un fluido es:

Paparente=Preal−Pfluido

Teniendo en cuenta que:

Preal = mcubo⋅gPfluido=E= dfluido⋅Vfluido⋅g

Como el cuerpo se sumerge completamente en el fluido, el volumen de fluido desalojado es exactamente el volumen del cubo. Por lo tanto si sustituimos los datos que nos proporcionan en el enunciado en la primera ecuación:

Paparente=mcubo⋅g−dfluido⋅Vfluido⋅g ⇒Paparente=10 kg ⋅9.8 m/s2 − 1000 kg/m3 ⋅10−3 m ⋅9.8 m/s2 ⇒Paparente = 88.2 N

7 0
2 years ago
In a heat engine if 1000 j of heat enters the system the piston does 500 j of work, what is the final internal energy of the sys
nydimaria [60]

Answer : The final energy of the system if the initial energy was 2000 J is, 3500 J

Solution :

(1) The equation used is,

\Delta U=q+w\\\\U_{final}-U_{initial}=q+w

where,

U_{final} = final internal energy

U_{initial} = initial internal energy

q = heat energy

w = work done

(2) The known variables are, q, w and U_{initial}

initial internal energy = U_{initial} = 2000 J

heat energy = q = 1000 J

work done = w = 500 J

(3) Now plug the numbers into the equation, we get

U_{final}-(2000J)=(1000J)+(500J)

(4) By solving the terms, we get

U_{final}-(2000J)=(1000J)+(500J)

U_{final}-(2000J)=1500J

U_{final}=2000J+1500J

U_{final}=3500J

(5) Therefore, the final energy of the system if the initial energy was 2000 J is, 3500 J

5 0
3 years ago
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