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patriot [66]
3 years ago
12

HELPPPPP !!!!

Physics
1 answer:
Nataly [62]3 years ago
4 0

Your answer is C)

a)t=2.78 sec

b)R=835.03 m

c)

Explanation:

Given that

h= 38 m

u=300 m/s

here given that

The finally y=0

So

t=2.78 sec

The horizontal distance,R

R= u x t

R=300 x 2.78

R=835.03 m

The vertical component of velocity before the strike

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
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Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

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the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

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Answer:

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