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3 years ago

Change 6 revolutions per minute to a linear speed measured in m/s. The radius is 3 m.

Physics

1 answer:

satela [25.4K]3 years ago

7 0

Answer:

Explanation:you know

circumference of a circle =2*pi*r=?

r=radius of a circle =3m

total distance of 6 revolution =s=6*2*pi*r

again, linear speed =v=s/t=?

t=time =60sec

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3 years ago

Why is it necessary for cells to be so small

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3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

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3 years ago

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4 years ago

Angular velocity in the z direction of a flywheel is w(t)=A + Bt2 The numerical values of the constants are A=2.75 and B=1.50. W

Ivanshal [37]

Answer:

α(0) = 0 rad/s²

α(5) = 15 rad/s²

Explanation:

The angular velocity of the flywheel is given as follows:

w(t) = A + B t²

where, A and B are constants.

Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:

Angular Acceleration = α (t) = dw/dt

α(t) = (d/dt)(A + B t²)

α(t) = 2 B t

where,

B = 1.5

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α(0) = 2(1.5)(0)

α(0) = 0 rad/s²

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α(5) = 2(1.5)(5)

α(5) = 15 rad/s²

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3 years ago