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vodka [1.7K]
4 years ago
6

Young David who slew Goliath experimented with slings beforetackling the giant. He found that he could revolve a sling oflength

0.600 m at the rate of9.00 rev/s. If he increased the lengthto 0.900 m, he could revolve the slingonly 7.00 times per second.
(a) What is the speed of the stone for eachrate of rotation?
1 m/s at 9.00 rev/s
2 m/s at 7.00 rev/s

(b) What is the centripetal acceleration of the stone at9.00 rev/s?
3 m/s2

(c) What is the centripetal acceleration at 7.00 rev/s?
4m/s2
Physics
1 answer:
Mars2501 [29]4 years ago
3 0

Answer:

a) v1 = 33.9 m/s , v2 = 39.6 m/s, b) a = 1015.4 m/s² and c)  a = 1742.4 m/s²

Explanation:

To solve this exercise we will use the relationship between angular and linear quantities

a) The angular and linear velocity are related by

     v = w r

Let's reduce the quantities to the SI system

    w1 = 9.00 rev / s (2π rad / 1rev) = 18 π rad / s = 56.55 rad / s

    w2 = 7.00 rev / s = 14π rad / s = 43.98vrad / s

     v1 = 56.55 0.6

     v1 = 33.9 m / s

   

     v2 = 43.98 0.9

    v2 = 39.6 m / s

b) centripetal acceleration for v1

     a = v² / r

     a = 33.9² / 0.6

     a = 1015.4 m / s²

c) centripetal acceleration for v2

     a = 39.6² / 0.9

     a = 1742.4 m / s²

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_____ you remember to put the lid back on the jar of mayonnaise​
Lilit [14]

Answer:

Did you remember to put the lid back on the jar of mayonnaise?

Explanation: Hope this helps :)

7 0
3 years ago
A portable basketball set has a base and a post arrangement. The post arrangement consists of a post, backboard, hoop and net. T
Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

  a) The minimum force of the wind that turns the system is Fw = 17.64 N

  b) The system resists much greater forces because the base has more mass

 Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

               Σ τ = 0

Where τ is the torque  

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

         cos 15 = \frac{y_w}{2.35}

         y_w = 2.35 cos 15

Post Weight

        sin 15 = \frac{x_p}{2.00}

         xp = 2.0 sin 15

Base weight

         cos (90-15) = \frac{x_b}{0.25}

         xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

     

          F_w \ y_w  + W_p \ x_p - W_b \ x_b = 0

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is W_p = 26.0 N and the weight of the base with water is W_b = 810 N

     F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}

     F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}

       

Let's  calculate

     F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}

     F_w = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

  a) the minimum force of the wind that turns the system is Fw = 17.64 N

  b) the system resists much greater forces because the base has more mass

Learn more  here: brainly.com/question/7031958

3 0
3 years ago
What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t
vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0
3 years ago
In what way is a screw similar to an inclined plane?
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They both have a mechanical advantage meaning that the force applied to the mechanism is not the same as the force produced.
5 0
3 years ago
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Why would the measurements have to take place around the same time every day
bixtya [17]
It is vital to take measurements at the same time everyday to determine the natural cycle, this will show the numbers on a regular basis. It is important because this will be basis of any action that needs to be taken. If it's not taken at the same time, it may disrupt the correct result or change the expected outcome.
5 0
3 years ago
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