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ANTONII [103]
3 years ago
12

Are lions stronger then tigers?

Physics
1 answer:
Oliga [24]3 years ago
6 0

Answer:

I would think no, because a tiger has a more difficult truain to live in while the lion lives in a flat plain, also lions are a pack animal, only the female lions do the hunting, thus making the lions not as active all of the time, where as the tiger is a solitary animal and does all of the hunting by it's self.

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Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let \alpha be the angular acceleration of the end of the rod

Taking torque about the link, we have:

\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

\tau = I_{rod}\ \alpha......(ii)

From equations (i) and (ii) we have:

mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }

The acceleration of the end of the rod farthest from the link is given by:

a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}

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Answer:

Answer 11

Explanation:

. Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them

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Velocity time graph work sheet what was the greatest velocity
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Can you give us a picture or something please!!

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3 years ago
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift i
lys-0071 [83]

Answer:

   W₃ = 3310.49 J ,  W3 = 3310.49 J

Explanation:

We can solve this exercise in parts, the first with acceleration, the second with constant speed and the third with deceleration. Therefore it is work we calculate it in these three sections

We start with the part with acceleration, the distance traveled is y = 5.90 m and the final speed is v = 2.30 m / s. Let's calculate the acceleration with kinematics

       v2 = v₀² + 2 a₁ y

as they rest part of the rest the ricial speed is zero

        v² = 2 a₁ y

        a₁ = v² / 2y

        a₁ = 2.3² / (2 5.90)

        a₁ = 0.448 m / s²

with this acceleration we can calculate the applied force, using Newton's second law

         F -W = m a₁

         F = m a₁ + m g

         F = m (a₁ + g)

         F = 69 (0.448 + 9.8)

         F = 707.1 N

Work is defined by

         W₁ = F.y = F and cos tea

As the force lifts the man, this and the displacement are parallel, therefore the angle is zero

          W₁ = 707.1 5.9

           W₁ = 4171.89 J  W3 = 3310.49 J

Let's calculate for the second part

the speed is constant, therefore they relate it to zero

           F - W = 0

           F = W

           F = m g

           F = 60 9.8

           F = 588 A

the job is

           W² = 588 5.9

            W2 = 3469.2 J

finally the third part

in this case the initial speed is 2.3 m / s and the final speed is zero

           v² = v₀² + 2 a₂ y

            0 = vo2₀² + 2 a₂ y

            a₂ = -v₀² / 2 y

            a₂ = - 2.3²/2 5.9

            a2 = - 0.448 m / s²

we calculate the force

            F - W = m a₂

            F = m (g + a₂)

            F = 60 (9.8 - 0.448)

            F = 561.1 N

we calculate the work

            W3 = F and

            W3 = 561.1 5.9

          W3 = 3310.49 J

total work

          W_total = W1 + W2 + W3

          W_total = 4171.89 +3469.2 + 3310.49

           w_total = 10951.58 J

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Answer:

You can describe the motion of an object by its position, speed, direction, and acceleration. An object is moving if its position relative to a fixed point is changing. idk if this helps.

Explanation:  

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