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Kisachek [45]
3 years ago
9

Two upwards forces act: one of 34N which is 3.5m to the left of the axle, the other is 25N and is 2.6m to the right of the axle.

There is also a 10N force to the left which is 0.80m below the axle.
Physics
1 answer:
KiRa [710]3 years ago
5 0

Answer:

See explanation below

Explanation:

In this case, you want to know if you put an object between these forces, which direction would go.

To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:

M = F * d   (1)

And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:

anticlockwise moment = clockwise moment.

The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.

Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns

Anticlockwise moment = 34 * 3.5 = 119 Ns.

As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.

Hope this helps

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Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
To aid in the prevention of tooth decay, it is recommended that drinking water contain 0.800 ppm fluoride, F−. How many grams of
Leviafan [203]

Answer:

2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-

Explanation:

Here are the steps of how to arrive at the answer:

The volume of a cylinder = ((pi)D²/4) × H

Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m

H = Height of the reservoir = 87.32m

Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³

1ppm = 1g/m³

0.8ppm = 0.8 × 1g/m³

= 0.8g/m³

Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.

Thank you for reading.

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use the pendulum equation to calcuate the period of a 1.50 pendulum. Remeber that the vaule of "g" is 9.81 m/s² please help ​
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Answer:

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square both sides to get 2.25 = 2\pi ( L/9.81)

multiply both sides by 9.81 then divide by 2 and 3.14 as a substitue for pi. The answer should be about 3.51 in length

L = 3.51

If this helps, mark me brainliest pls

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