lomework 3 Begin Date:9/7/2018 12:01:00 AMDue Date: 9/13/2018 11:59.00 PM End Da

Question

lomework 3 Begin Date:9/7/2018 12:01:00 AMDue Date: 9/13/2018 11:59.00 PM End Date:9/13/2018 11:59.00 PMM (6%) Problem 3: of t # 0.29 s. A boxer's fist and glove have a mass of m-0.86 kg. The boxer's fist can obtain a speed of v-5.75 ms in a time @theexperta com . tricking id: 8075-BE-5C47-B250 18965 In accordance with Erpert TAS Terms of Service pying this information to any solutions shar websie is strictly fortiddes Doing so may resalt in seination f your Expen TA Account g 50% Part (a) Find the magnitude of the average acceleration aave in meters per square second of the boxer's fist. Grade Suns Dedactions sino I coso I tan() | |( 7| 8|9 4 5 6 Amempts ren (04 per atter cotanO asin)0 acosO atano acotanO sinh0 os tanh0 cotanhO O Degrees Radians I give up Hints: Feedback: 0% deduction per foedback. d 50% Part (b) How much force did the boxer apply to his fisuglove, in newtons? do 4

Explanation / Answer

Solution :-

                  Given:-

                             Speed (v) = 5.75 m/s

                             Mass (m) = 0.86 kg

                              Time (t) = 0.29 s

The formula for magnitude of average acceleration is given by

Aav = speed / time

Aav = V / t

Where ,

V = velocity or speed in meter

T = time in second

              Aav = 5.75 m/s / 0.29 s

Aav = 19.28 m /s2

The magnitude of average acceleration is = 19.82 m/s2

The formula for force is given by

According to newtons second law the formula for force is given by

F = ma

where ,

   F = Force in newton

m = mass in kilogram

a = acceleration in meter / second square

F =0.86 kg / 19.82 m/s2

                   F = 0.043 newton

The force is given by 0.043 newton

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