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Mariana [72]
3 years ago
7

A girl standing upright exerts a pressure of 15000 N/m2 on the floor. Given that the total area of contact of shoes and the floo

r is 0.02m2 Determine the pressure she would exert on the floor if she stood on one leg
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

Answer:

30,000N/m²

Explanation:

Given parameters:

Pressure on two legs = 15000N/m²

Total surface area of contact = 0.02m²

Surface area on one leg = \frac{0.02}{2}  = 0.01m²

Unknown:

Pressure exerted if she stood on one leg = ?

Solution:

Pressure is the force per unit area on a body.

      Pressure  = \frac{force}{area}

 The force she is exerting is due to the weight her both. Standing on both feet and on a single feet is based on the same source of force;

         Force  = pressure x area

         Force = 15000 x 0.02  = 300N

So;

  Pressure from one leg = \frac{300}{0.01}   = 30,000N/m²

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This type of star is called the "white dwarf." When a very massive star exhausts its nuclear fuel it explodes as a supernova.

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Two sacks contain the same number of identical apples and are separated by a distance r. The two
andrezito [222]

Answer:

The appropriate response will be "F_2=\frac{3}{4}F". A further explanation is given below.

Explanation:

According to the question,

⇒ F_1=\frac{G(m_1 m_2)}{r^2} ....(equation 1)

and,

⇒ F_2=\frac{G(m_1 m_2)}{r^2} ....(equation 2)

Now,

On dividing both the equations, we get

⇒ \frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}

⇒ \frac{F_1}{F_2}= \frac{4}{3}

⇒ F_2=\frac{3}{4}F

4 0
3 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

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What are Base Quantities ?​
Norma-Jean [14]

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<em>physical quantities that cannot be defined in terms of other quantities.</em>

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A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the
My name is Ann [436]

Answer:

2.82 s

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Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

7 0
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