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Svet_ta [14]
3 years ago
7

Tammy is a forensic investigator examining a body at a crime scene. She notes that the body is stiff but still flexible. What do

es she know about the time of death?
Physics
1 answer:
wlad13 [49]3 years ago
8 0
I may be wrong but does it mean it was revent? because i know shortly after someone dies your body becomes fully stiff so maybe it was recent and it's in the process off stiffening up
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Assume your mass is 60 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb
Andre45 [30]

Answer:

W=1705.2 J

Explanation:

Given that

mass ,m= 60 kg

Acceleration due to gravity ,g= 9.8 m/s²

Height ,h= 2.9 m

As we know that work done by a force given as

W = F . d

F=force

d=Displacement

W=work done by force

Now by putting the values

F= m g (Acting downward  )

d= h  (Upward)

W= m g h    ( work done against the force)

W= 60 x 9.8 x 2.9 J

W=1705.2 J

Therefore the answer will be 1705.2 J.

8 0
3 years ago
Section 1: Experimental Overview
Luden [163]

Answer:

1-state what the lab is about, that is, what scientific concept (theory, principle, procedure, etc.) you are supposed to be learning about by doing the lab. You should do this briefly, in a sentence or two. If you are having trouble writing the opening sentence of the report, you can try something like: "This laboratory experiment focuses on X…"; "This lab is designed to help students learn about, observe, or investigate, X…." Or begin with a definition of the scientific concept: "X is a theory that…."

2-give the necessary background for the scientific concept by telling what you know about it (the main references you can use are the lab manual, the textbook, lecture notes, and other sources recommended by the lab manual or lab instructor; in more advanced labs you may also be expected to cite the findings of previous scientific studies related to the lab). In relatively simple labs you can do this in a paragraph following the initial statement of the learning context. But in more complex labs, the background may require more paragraphs.

Explanation:

7 0
3 years ago
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t
Marysya12 [62]
From the law of conservation of momentum
m1u1+ m2u2= m1v1+ m2v2
110*8+ 110*-10= 110*-10 + 110* v2
v2= 8 m/sec
8 0
3 years ago
An ant travels 2.78 cm (West) and then turns and travels 6.25 cm (South 40 degrees East). What is the ant's total displacement?
andrey2020 [161]
From\ cosine\ theorem:\\\\
c^2=a^2+b^2-2abcos(\angle between\ a\ and\ b)\\\\
a=2,78\\b=6,25\\
\angle between\ a\ and\ b=90+50=140^{\circ}\\\\
c^2=2,78^2+6,25^2-2*2,78*6,25cos(140^{\circ})\\\\
c^2=7,7284+39,0625-34,75*(-0,77)\\\\c^2=46,7909+26,7575\\\\c^2=735484\\\\c=8,58cm\\\\Total\ displacement\ is\ equal\ to\ 8,58cm.

5 0
3 years ago
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