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grigory [225]
3 years ago
10

I need the answer to number #23 please will give BRAINLIEST

Physics
1 answer:
Murrr4er [49]3 years ago
3 0
Yeah lowkey i think it’s B
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Determine the amount of torque
malfutka [58]

Answer:

250Nm

Explanation:

Given parameters:

Length of the long pry bar  = 1m

Force acting on it  = 250N

Angle  = 90°

Unknown:

Amount of torque applied  = ?

Solution:

Torque is the turning force on a body that causes the rotation of the body.

The formula is given as:

 Torque  = Force x r Sin Ф  

r is the distance

 So;

   Torque  = 250 x 1 x sin 90  = 250Nm

7 0
3 years ago
Which of the following equations is balanced correctly? A. 3 H2O → H2 + 3 O2 B. Cl2 + 2 KBr → KCl + Br2 C. 2 C2H2 + 5 O2 → 4 CO2
posledela
A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.
8 0
3 years ago
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Find the equilibrant of two 10.0-N forces acting upon a body when the angle between the forces is 90° Solve graphically using a
bazaltina [42]

The equilibrant force of the two given forces is 14.14 N.

<h3 /><h3 /><h3>What is equilibrant force?</h3>
  • This is a single force that balances other given forces.

The given parameters:

  • First force, F₁ = 10 N
  • Second force, F₂ = 10 N
  • Angle between the forces, θ = 90⁰

The equilibrant force of the two given forces is calculated as follows;

F = \sqrt{F_1 ^2 + F_2 ^2} \\\\F = \sqrt{(10)^2 + (10)^2} \\\\F = 14.14 \ N

Thus, the equilibrant force of the two given forces is 14.14 N.

Learn more about equilibrant force here: brainly.com/question/8045102

5 0
2 years ago
I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

3 0
3 years ago
This whole worksheet. I am not good at this and I am completely lost
Ket [755]
1.) potential energy
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4.)potential energy
5.)kinetic energy
6.) potential and kinetic energy
3 0
3 years ago
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