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Anarel [89]
3 years ago
8

2in+3in+4in+4in+6in+7in calculate the area and perimeter of each shape

Physics
2 answers:
azamat3 years ago
5 0

Answer:

Let "a", "b" and "с"  be sides of the triangle ("с" is the longest side).

The triangle will be:

right if       a² + b² = c²

аcute if     a² + b² > c²    

obtuse if   a² + b² < c²    

a.

a=3, b=4 and c=5

a² + b² = 3² + 4² = 9 + 16 = 25   and   c² = 5² = 25

25 = 25   ⇒  right triangle.

b.

a=5, b=6 and c=7

a² + b² = 5² + 6² = 25 + 36 = 61   and   c² = 7² = 49

61 > 49   ⇒  аcute triangle.

c.

a=8, b=9 and c=12

a² + b² = 8² + 9² = 64 + 81 = 145   and   c² = 12² = 144

145 > 144   ⇒  аcute triangle.

ololo11 [35]3 years ago
4 0
Perimeter=26in

Area would be to multiply the units together
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Answer:

Velocity = 4.33[m/s]

Explanation:

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All this energy will become kinetic energy and we can find the velocity.

37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]

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This means the lungs are inflated with - Highly pressurized- gas.

This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.

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If the acceleration of motorboat is 4m/s^2 and the motorboat stsrtsfrol.Rest what is velocity after 6.0 s
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Answer:

24 m/s

Explanation:

Using v = u + at where u = initial velocity of the motorboat = 0 m/s (since the boat starts from rest), a = acceleration = 4 m/s², t = time = 6 s and v = velocity of the motorboat after 6.0 s.

Substituting the values of the variables into the equation, we have

v = u + at

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Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite
nalin [4]

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s

Hence, this is the required solution.

7 0
3 years ago
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