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2 years ago

All of the following could be considered reconstruction except

1 answer:

8 0

The first three choices: a, b and c can be considered reconstruction except the last one which is letter d. I'm not really certain what reconstruction is, but judging from the patterns of the first three choices, reconstruction could mean that an inference is made after a part of an event has proved that event to be true.

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If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/

Goshia [24]

If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)

this question is troubling me i guessed 96 ft/s

can someone help me out and explain it thanks so much!!!!!!

7 0

2 years ago

Read 2 more answers

There is a force which resists changes in motion, called

adoni [48]

Answer:

inertia 11

Explanation:

5 0

2 years ago

n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

2 years ago

How much work would it take to push two protons very slowly from a separation of 2.00×10−10m (a typical atomic distance) to 3.00

laiz [17]

Answer:

Work= -7.68×10⁻¹⁴J

Explanation:

Given data

q₁=q₂=1.6×10⁻¹⁹C

r₁=2.00×10⁻¹⁰m

r₂=3.00×10⁻¹⁵m

To find

Work

Solution

The work done on the charge is equal to difference in potential energy

W=ΔU

$Work=U_{1}-U_{2}\\ Work=-kq_{1}q_{2}[\frac{1}{r_{2}}-\frac{1}{r_{1}} ]\\Work=(-9*10^{9})*(1.6*10^{-19} )^{2}[\frac{1}{3.0*10^{-15} }-\frac{1}{2*10^{-10} } ]\\ Work=-7.68*10^{-14}J$

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3 years ago

Calculate the weight of your energy car if it has a mass of 233.85 kg.

lianna [129]

Explanation:

'gerfe4egd4gwdrferdfw14589u99y999q EP and it's not the same thing to say about this is a good day for the next few weeks of age or something else I would have to do it for you and your friends to be the first to see the new album and I am a big deal to the world is not the same thing to say about this one for you and I am a big deal to me that he was the first to see you in a few years back and forth to see the world is not a fan and a few years back and I have to be the best way possible for the next few days later and the other side is the only thing that you can get it right here is the only thing I can get it right here is a great time to get a free download of my life

3 0

1 year ago