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melomori [17]
3 years ago
7

¿Cuál es la formula para resolver la caída libre?

Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Caída Libre

h= v0 t+1/2 g t2

vf = v0 + g t

?

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A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
If the man has initial velocity of 0.4 m/s, acceleration is 0.343 m/s^2, a time of 5.8 seconds, and experimental final velocity
agasfer [191]

Answer:

The answer is going to be C.

Explanation:

Trust me. Im an expert in physics

6 0
3 years ago
What is the mass of an object that weighs 500 newtons on earth?
Eva8 [605]
The mass would be 51 kg. So b is the right answer.
6 0
4 years ago
A piece of wood on top of an ocean wave stays in the same location, only moving up and down as the wave passes. It is energy tha
Ierofanga [76]

Answer:

Hey

Yes, this is true.

As some people have it wrong, waves in the water (ocean) are not waves of moving water, rather the wave is moving through the water. A wave is a disturbance of a medium not the meduim moving.

8 0
3 years ago
Determine the value of the resultant and its location from O.<br>see attach image.​
xxTIMURxx [149]

Answer:

Explanation:

In the x direction the force will be

½(-w₀)L/2 = -¼w₀L  

acting ⅔(L/2) = L/3 below the x axis.

In the y direction the force will be

½(-w₀)L + ½w₀L/2 = -¼w₀L  

the magnitude of the resultant will be

F = w₀L  √((-¼)² + (-¼)²) = w₀L√⅛

in the direction

θ = arctan(-¼w₀L / -¼w₀L) = 225°

to find the distance, we balance moments

(w₀L√⅛)[d] = ½(w₀)L[⅔L] + ¼w₀L[⅔L/2] - ¼w₀L[L - ⅓L/2]

     (√⅛)[d] = ½         [⅔L] + ¼      [⅔L/2] - ¼      [L - ⅓L/2]

     (√⅛)[d] = ½[⅔L] + ¼[⅔L/2] - ¼[L - ⅓L/2]

     (√⅛)[d] =      ⅓L  +    ⅟₁₂L     -  ¼L + ⅟₂₄L  

     (√⅛)[d] = 5L/24

               d = 5L/24 / (√⅛)

               d = 5√⅛L/3

8 0
3 years ago
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