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Setler [38]
3 years ago
5

An irregular object of mass 3 kg rotates about an axis, about which it has a radius of gyration of 0.2 m, with an angular accele

ration of 0.5 rad.s?. The magnitude of the applied torque is: a) 0.30 N.m b) 3.0 x 102 N.m C) 0.15 N.m d) 7.5 x 102 N.m e) 6.0 x 102 N.m.
Physics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

0.06 Nm

Explanation:

mass of object, m = 3 kg

radius of gyration, k = 0.2 m

angular acceleration, α = 0.5 rad/s^2

Moment of inertia of the object

I = mK^{2}

I = 3 x 0.2 x 0.2 = 0.12 kg m^2

The relaton between the torque and teh moment off inertia is

τ = I α

Wheree, τ is torque and α be the angular acceleration and I be the moemnt of inertia

τ = 0.12 x 0.5 = 0.06 Nm

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To find some of the parameters characterizing an object moving in a circular orbit.The motivation for Isaac Newton to discover h
tamaranim1 [39]

Answer:

K = \frac{GMm}{2r}

T^2 = 4\pi^2(\frac{r^3}{GM})

Explanation:

As we know that for a satellite the force of gravitation is equal to the centripetal force

so we will have

F = \frac{GMm}{r^2}

\frac{mv^2}{r} = \frac{GMm}{r^2}

so we know that kinetic energy is given as

K = \frac{1}{2}mv^2

so we have

K = \frac{GMm}{2r}

now for time period we know

T = \frac{2\pi r}{v}

from above expression of kinetic energy we have

v = \sqrt{\frac{GM}{r}}

so we have

T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}}

so square of time period is given as

T^2 = 4\pi^2(\frac{r^3}{GM})

3 0
3 years ago
he drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 7 scale model. The tunnel operates with fre
dusya [7]

Answer:20.03 m/s

Explanation:

Given

L_r=1:7

velocity of Prototype v_p=53 m/s

Taking Froude number same for both flow as it is a dimensionless number for different flow regimes in open Flow

(\frac{v_m}{\sqrt{L_mg}})=(\frac{v_p}{\sqrt{L_pg}})

v_m=v_p\times \sqrt{\frac{L_m}{L_p}}

v_m=53\times \frac{1}{\sqrt{7}}

v_m=20.03 m/s

           

4 0
2 years ago
A Foucault pendulum consists of a brass sphere with a diameter of 32.0 cm suspended from a steel cable 10.0 m long (both measure
Sergio039 [100]

Answer:

T = 44.35 °C

Explanation:

d = 32cm

R = 16 cm

Lsteel = 10m

T1 = 20° C

Space = 0.3cm

The space between the sphere and the floor is represented by δL(total) after the temperature increases.

As the temperature increases, both will expand.

So,

0.3 x 10^(-2) = δL(steel) + δR(brass)

= {L(o) x α(steel) x δT} + {R(o) x {α(brass) x δT}

= {10 x 1.2 x 10^(-5) x (T-20)} + {0.16 x 2 x 10^(-5) x (T-20)}

= 12.32 x 10^(-5) x(T-20)

Therefore (T-20) = (0.3 x 10^(-2)) / {12.32 x 10^(-5)}

T = 20 + 24. 35 = 44.35 °C

8 0
2 years ago
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