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larisa86 [58]
3 years ago
15

How do you determine the average speed of an object in cm/s

Physics
1 answer:
vaieri [72.5K]3 years ago
6 0

Answer:

Average speed can be calculated by dividing the total distance the object travels by the total amount of time it takes to travel that distance.

Explanation:

Hope you understand.

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How does an octopus or a squid uses Newton's third law of motion to escape from enemies
Sloan [31]
I think everytime they swim away, the water pushes the enemy back, which makes the octopus go faster. Every action has an equal and opposite reaction. Hope this helps.
4 0
4 years ago
What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem
8090 [49]

Answer:

\Delta L = 4.2 m

Explanation:

As per the formula of thermal expansion we know that

L = L_o(1 + \alpha\Delta T)

so here we will have

L_o = 1400 m

\alpha = 12 \times 10^{-6} per ^oC

\Delta T = 250 degree C

so here change in the length of the rod is given as

\Delta L = L - L_o

\Delta L = L_o \alpha \Delta T

\Delta L = 1400 (12 \times 10^{-6})(250)

\Delta L = 4.2 m

8 0
4 years ago
If you just saw a shooting star and wanted to write about the moment, which would be the best type of poem to write? a haiku a n
Norma-Jean [14]
I personally would write it in a Ballad format.
Ballad poems tell a story.

Narrative poems tells a story often making the voices of the/a narrator or character.

Haiku poems rarely rhyme. 

Sonnet poems are rhythmic.
6 0
4 years ago
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
3 years ago
In a pig-calling contest, a caller produces a sound with an intensity level of 60 db. how many such callers would be required to
zloy xaker [14]

The imaginary second level is 60 dB more intense than the real level of the caller.

60 dB means a multiplication of  10⁶  = <em>1 million.</em>

That's how many equally-talented callers it would take to be 60 dB louder than him.

7 0
3 years ago
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