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8_murik_8 [283]
3 years ago
8

I'm not quite sure how the formula works. I thought it was

Physics
1 answer:
juin [17]3 years ago
4 0

Answer:

See the explanation below.

Explanation:

Of course yes, after my explanation you will understand very well how to analyze this type of problems.

First, we must differentiate the weight of the mass, the mass has units of kilograms or pounds, while the weight is the product of the mass by gravitational acceleration. In this way, we have the following formula for the weight of anybody.

w = m*g

where:

w = weight [N] (units of Newtons]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

When we have the product of these units [kg]*[m/s²], We are always going to get as a result the units of Newtons [N], The Newton is a unit of force.

Now, to solve this problem we must use Newton's second law, which tells us that the sum of forces on a body must be equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF  = forces acting over the gymnast [N]

m = mass = 55 [kg]

a = desacceleration = 6*9.81 = 58.86 [m/²]

Now we have two forces acting over the gymnast the first one is his or her weight (the mass by the gravity acceleration) and the second force is the one made it by his or her legs upwards in order to be standing. Let's take as positive signs the movements and forces upwards and negative downwards

- (m*g) + Fl = m*a

where:

Fl = force made it by the legs [N]

As the deceleration is upward because it prevents the person from continuing to go down, it is taken as a positive sign.

- (55*9.81) + Fl = (55*58.86)

Fl =  (55*58.86) + (55*9.81)

Fl = 3776.85 [N]

I hope you can understand :)

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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

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t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

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Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

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a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

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Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

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3 years ago
PLEASE HELP
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Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

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0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m

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Answer:

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