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Ilia_Sergeevich [38]
3 years ago
11

Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8

kg, 31.5 kg, and 25.9 kg, respectively.
What is the new velocity of the sled if Kid A jumps off vertically from the sled? (Kid A velocity = 0)Three kids are riding on a snow sled traveling horizontally without friction at 19.8 m/s. The masses of Kid A, B, and C are 42.8 kg, 31.5 kg, and 25.9 kg, respectively.

What is the new velocity of the sled if Kid A jumps off vertically from the sled? (Kid A velocity = 0)
Physics
1 answer:
Kamila [148]3 years ago
4 0

Answer:

34.6 m/s

Explanation:

From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg

Final mass will be 31.5+25.9=57.4 kg

From formula of momentum

M1v1=m2v2

Making v2 the subject of the formula then

V2=\frac {M1v1}{m2}

Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then

V2=\frac {100.2 kg\times 19.8 m/s}{57.4 kg}=34.56376 m/s\approx 34.6 m/s

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A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate t
vivado [14]

Answer:

<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction  opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move  can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

          μ is the coefficient of friction ( μs, in this case, static friction);

          m  is mass of the object and;

          g is the acceleration due to gravity( a constant equal to 9.81 m/s^{2})

from the equation we are provide with;

       μs  = 0.25

       m = 50 kg

       g =  9.81 m/s^{2}

      F =?

Using equation 1

F = 0.25 x 50 kg x  9.81 m/s^{2}

F = 122.63 N  

<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>

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The mass number of a nucleus (except Hydrogen) is:
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In the case of Hydrogen it depends of isotope of Hydrogen . 
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A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

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