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Anvisha [2.4K]
3 years ago
11

Describe an example of newton’s 3rd law of motion

Physics
1 answer:
professor190 [17]3 years ago
8 0

Answer:

For example, when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air. Engineers apply Newton's third law when designing rockets and other projectile devices.

Explanation:

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A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

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Are any rocks that form from another rock as a result of change in blank or blank
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Rotational inertia decreases proportional to the decrease in the radius of rotation.

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How many cups of water should you drink in a day?
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about 2.7liters for women and 3.7liters for men

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2 years ago
A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the
ser-zykov [4K]

Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

                       = 8.17 N/m

The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

5 0
3 years ago
Read 2 more answers
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