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xz_007 [3.2K]
3 years ago
5

Suppose the initial kinetic energy and final potential energy in an experiment are both zero. What can you conclude?

Physics
2 answers:
faltersainse [42]3 years ago
4 0
At the beginning it is not moving and at the end it is on the ground or the bottom of a hill.
puteri [66]3 years ago
3 0

Answer:

That an item is neither moving nor staying still in a position that is building up energy.

Explanation:

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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
2 years ago
Mike says, I gave up guitar because I’ll never be good at it, some people seem to have a good ear music, not me. Which mindset i
AveGali [126]
A negative mind set
8 0
2 years ago
An unknown galaxy has a large flattened core. Which of the following classifications would best fit this galaxy's description? I
Alja [10]

Answer:

Spiral

i know it is so dont say nun people

Explanation:

7 0
3 years ago
Read 2 more answers
A. <br> B.<br> C.<br> D.<br> Please help
MA_775_DIABLO [31]

Answer:

The answer is A, B, C and D

Explanation:

(is that how it works?)

8 0
2 years ago
Read 2 more answers
What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}

E(0.89) = 306500 N/C

3 0
3 years ago
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