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4 years ago

Need answers fast plz

Physics

1 answer:

Len [333]4 years ago

5 0

Answer:

Step-by-step explanation:

We have:

two right triangles with legs 4m and 6m

The area:

$A_1=\dfrac{1}{2}(4)(6)=12\ m^2$

three rectangles

22m × 6m

The area:

$A_2=22\cdot6=132\ m^2$

22m × 7.21m

The area:

$A_3=22\cdot7.21=158.62\ m^2$

22m × 4m

The area:

$A_4=22\cdot4=88\ m^2$

The Surface Area:

$S.A.=2A_1+A_2+A_3+A_4$

Substitute:

$S.A.=2(12)+132+158.62+88=402.62\ m^2$

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4 years ago

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4 years ago

A 60 g ball is dropped from rest from a height of 2.4 m. It bounces off the floor and rebounds to a maximum height of 1.9 m. If

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Answer:

The force is 1.34 newtons and its direction is upward.

Explanation:

Choosing positive direction pointing towards the floor in this collision we're going to use the momentum-impulse theorem that states:

$J=\Delta p$ (1)

with $\Delta p=p_f-p_i$ the change in the momentum and J the impulse, with pi the initial momentum that is the momentum just before the collision and pf the final momentum th is the momentum just after the collision. The impulse J is also defined as:

$J=F_{avg}\Delta t$ (2)

with $F_{avg}$ the average force and $\Delta t$ the time the collision lasts

We can equate expressions (2) and (1):

$\Delta p=p_f-p_i=F_{avg}\Delta t$

Using the definition of linear momentum as mass (m) time velocity (v):

$mv_f-mv_i=F_{avg}\Delta t$

We can solve for Favg:

$F_{avg}=\frac{m(v_f-v_i)}{\Delta t}$ (3)

Now we should find the velocities vf and vi, we should do this using conservation of energy:

For the velocity the ball has just before reaches the floor:

$U_i=K_f$

With Ui the initial potential energy (there is not initial kinetic energy) and Kf the final kinetic energy (there is not final potential energy), then:

$mgh=\frac{mv_i^2}{2}$

solving for vi:

$v_i=\sqrt{2gh}=\sqrt{2*9.81*2.4}=6.86\frac{m}{s}$

For the velocity the ball has just after bounces the floor:

$K_i=U_f$

There is not initial potential energy because it's a floor level at this instant, and the there is not final kinetic energy because the ball has instantly zero velocity at its maximum height (hm), then:

$\frac{mf_i^2}{2}=mgh_m$

solvig for vf:

$v_f=\sqrt{2gh_m}=\sqrt{2*9.81*1.9}=6.10\frac{m}{s}$

Using vf and vi on (3):

$F_{avg}=\frac{(0.06)(6.10-6.86)}{0.034}=-1.34 N$

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