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3 years ago

What conditions must be present for a translational and rotational equilibrium of a rigid body?

1 answer:

5 0

If a rigid body is in translational and rotational equilibrium, then
1. The sum of all applied forces is zero.
2. The sum of all applied moments is zero.
3. There is no linear acceleration.
4. There is no rotational acceleration.

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4 years ago

A granite monument has a volume of 25,365.4 cm3. The density of granite is 2.7 g/cm3. Use this information to calculate the mass

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3 years ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 2.33 m away from a waterfall 0.488 m in heigh

Ulleksa [173]

Answer:

5.68 m/s

Explanation:

The motion of the salmon is the same as a projectile: it is launched with an initial speed $u$ at an angle of $\theta=38^{\circ}$ above the horizontal.

The motion of the salmon consists of two indipendent motion:

- Along the horizontal direction, it is a uniform motion with constant velocity

$v_x = u cos \theta$

So that the distance travelled is

$d=v_x t = u cos \theta t$ (1)

- Along the vertical direction, it is a uniformly accelerated motion with constant acceleration downward, so the vertical displacement is

$y = u sin \theta t - \frac{1}{2}gt^2$ (2)

where g is the acceleration of gravity.

We know the following:

- The horizontal distance travelled by the salmon to reach the waterfall is

d = 2.33 m

- The vertical distance travelled is the height of the waterfall,

y = 0.488 m

From (1) we get:

$t=\frac{d}{u cos \theta}$

And substituting into (2), we can solve the equation to find t, the time at which the salmon reaches the waterfall:

$y = u sin \theta \frac{d}{u cos \theta} - \frac{1}{2}gt^2 = d tan \theta - \frac{1}{2}gt^2\\t = \sqrt{\frac{2(d tan \theta - y)}{g}}=\sqrt{\frac{2(2.33 tan 38^{\circ}-0.488)}{9.81}}=0.521 s$

And then, we can use eq.(1) again to find the initial speed, u:

$u=\frac{d}{cos \theta t}=\frac{2.33}{cos(38^{\circ})(0.521)}=5.68 m/s$

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3 years ago