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Minchanka [31]
2 years ago
11

Please help !! ASAP

Physics
1 answer:
umka2103 [35]2 years ago
5 0

Answer:

22) 5.6 Kg.m/s

23) 0.067 Kg

24) 1.16 m/s

Explanation:

Momentum is found by multiplying mass and velocity

P=mv where P is momentum, m is mass of object and v is the velocity

22

Given

Mass, m=2 Kg

Rate, v=2.8 m/s

Relationship

P=mv

Solution

P=mv=2*2.8=5.6 Kg.m/s

23

Given

Momentum, p=0.1 Kg.m/s

Rate, v=1.5 m/s

Relationship

P=mv and making m the subject then

m=\frac {P}{v}

Solution

m=\frac {0.1 Kgm.s}{1.5 m/s}\approx 0.067 Kg

24

Given

Momentum, p=4.9\times 10^{8} Kg.m/s

Mass, m=4.23\times10^{8} Kg

Relationship

P=mv and making v the subject then

v=\frac {P}{m}

Solution

m=\frac {4.9\times 10^{8}  Kgm.s}{4.23\times10^{8} }\approx 1.16 m/s

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An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
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a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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3 years ago
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