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3 years ago

Answer my question plzz !!any body

Physics

1 answer:

Mice21 [21]3 years ago

4 0

Answer:

The height is 80 meters.

Explanation:

The equation for the distance in the second segment of the fall is as below. In it, we use the initial velocity v1 that the ball had after completing the first segment. From this equation, v1 can be determined:

$h_2 = \frac{1}{2}gt_2^2+v_1t_2=0.5\cdot 10\frac{m}{s^2}\cdot 4 s^2+v_1\cdot 2 s=60m\implies\\v_1 = 20\frac{m}{s}$

Next, we use the kinematic equation for velocity at the end of the first segment of a free fall, to determine h1:

$v_1^2 = 2gh_1+v_0^2 = 2gh_1\implies h_1 = \frac{v_1^2}{2g}=20m$

The total height is then

$h = h_1 + h_2 = 20m + 60 m = 80m$

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Hoochie [10]

Answer:

26b) 66.7%

27) 500 N

Explanation:

26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.

26.b) Mechanical advantage is the ratio between the forces:

MA = load force / effort force

Efficiency is the ratio between the work:

e = work done on load / work done by effort

Work is force times distance.

e = (F load × d load) / (F effort × d effort)

Rearranging:

e = (F load / F effort) × (d load / d effort)

e = MA × (d load / d effort)

In a two pulley system, the load moves half the distance of the effort. So the efficiency is:

e = (4/3) × (1/2)

e = 2/3

e = 66.7%

27) In a three pulley system, the load moves a third of the distance of the effort.

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3 years ago

provides some pertinent background for this problem. A pendulum is constructed from a thin, rigid, and uniform rod with a small

gavmur [86]

Answer:

the period of the physical pendulum is 0.498 s

Explanation:

Given the data in the question;

$T_{simple$ = 0.61 s

we know that, the relationship between T and angular frequency is;

T = 2π/ω ---------- let this be equation 1

Also, the angular frequency of physical pendulum is;

ω = √(mgL / $I$ ) ------ let this equation 2

where m is mass of pendulum, L is distance between axis of rotation and the center of gravity of rod and $I$ is moment of inertia of rod.

Now, moment of inertia of thin uniform rod D is;

$I$ = $\frac{1}{3}$ mD²

since we were not given the length of the rod but rather the period of the simple pendulum, lets combine this three equations.

we substitute equation 2 into equation 1

we have;

T = 2π/ω OR T = 2π/√(mgL/ $I$ ) OR T = 2π√( $I$ /mgL)

so we can use $I$ = $\frac{1}{3}$ mD² for moment of inertia of the rod

Since center of gravity of the uniform rod lies at the center of rod

so that L = $\frac{1}{2}$ D.

now, substituting these equations, the period becomes;

T = 2π/√( $I$ /mgL) OR T = $2\pi \sqrt{\frac{\frac{1}{3}mD^2 }{mg(\frac{1}{2})D } }$ OR T = 2π√(2D/3g ) ----- equation 3

length of rod D is still unknown, so from equation 1 and 2 ( period of pendulum ),

we have;

ω $_{simple$ = 2π/ $T_{simple$ OR ω $_{simple$ = √(g/D) OR ω $_{simple$ = 2π√( D/g )

so we simple solve for D/g and insert into equation 3

so we have;

T = √(2/3) × $T_{simple$

we substitute in value of $T_{simple$

T = √(2/3) × 0.61 s

T = 0.498 s

Therefore, the period of the physical pendulum is 0.498 s

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2 years ago