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Vera_Pavlovna [14]
3 years ago
10

Match the following moves to the correct total distance and displacement of the moves.

Physics
1 answer:
MAVERICK [17]3 years ago
5 0
A. Move 2 m east and then 12 m east; displacement is 14 m east and the distance is 14 m
B. Move 10 m east and then 12 m west, the displacement is 2 m west and the distance is 22 m.
C. Move 8 m west and then 16 m east; the displacement is 8 m east and the distance is 24 m
D. Move 12 m west and then 8 m east; the displacement is 4 m and the distance is 20 m
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A giant armadilo moving northward with a constant acceleration covers the distance between two points 60m apart in 6 seconds. It
Naddika [18.5K]
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. To determine acceleration, we need to know the initial velocity and the final velocity and the time elapsed. From the given values, we need t o calculate for the initial velocity. We use some kinematic equations. We do as follows:

 x = v0t + at^2/2
60 = v0(6) + a(6)^2/2
60 = 6v0 + 18a          (EQUATION 1)

vf = v0 + at
15 = v0 + a(6)
15 = v0 + 6a             (EQUATION 2)

Solving for v0 and a,
v0 = 5 m/s
a = 1.7 m/s^2
8 0
3 years ago
A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0
2 years ago
The acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. Th
Lorico [155]
1/3 the weight than it is on earth, duh
3 0
3 years ago
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a distant galaxy is studied with a radio telescope x ray telescope and optical light telescope. the images form which set of mus
aivan3 [116]

Answer:

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Explanation:

  • Scientists and research workers are in constant search for more answers as they explore the universe and implement the laws of physics on the celestial entities. But, most of the objects inside the universe are not visible to human naked eye, as they are far from sight and thus more advanced form of instruments like the x-ray, optical, and light telescopes are used to determine the different properties of the celestial entities inside the universe.
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WILL UPVOTE EVERY ANSWER! MULTIPLE CHOICE QUESTION!
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B hardness

Giddy UP!!!!!
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