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Y_Kistochka [10]
3 years ago
11

You push on a box with 100 N of force, causing it to accelerate at 5 m/s?.

Physics
2 answers:
Xelga [282]3 years ago
4 0
<h3><u>Given</u><u>:</u><u>-</u></h3>

Force,F = 100 N

Acceleration,a = 5 m/s²

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the mass of the box .

<h3><u>Formula</u><u> </u><u>used:-</u><u> </u></h3>

\bf \: Force = Mass  \times  Acceleration

<h3><u>Solution:-</u><u> </u></h3>

\sf \: Force = Mass × Acceleration

★ Substituting the values in the above formula,we get:

\sf \implies \: 100 = Mass \times 5

\sf \implies \: Mass =   \cancel\dfrac{100}{5}

\sf \implies \: Mass = 20 \: kg

Nezavi [6.7K]3 years ago
4 0

We are given with the amount of force applied and accleration of a certain body and we have to find the mass of the box.

  • Force applied = 100 N
  • Accleration = 5 m/s²

<u>According to Netwon's 2nd law,</u>

After deriving, we will get that

  • Force = Mass × Accleration

Plugging the given values to get mass,

➝ 100 N = Mass × 5 m/s²

➝ Mass = 100 / 5 kg

➝ Mass = 20 kg

Thus, the required mass of the box is 20 kg.

Carry On Learning !

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Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1
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Answer:

\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.

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<u>Given:</u>

  • Mass, \rm m\pm\Delta m = 1.3\pm 0.4\ kg.
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\rm \Delta m,\ \Delta v are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.

The uncertainty in kinetic energy is given as:

\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.

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1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


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