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mixas84 [53]
3 years ago
7

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Physics
1 answer:
Paha777 [63]3 years ago
4 0

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

P_1=P_2

0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2

v_2=1 m/s

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

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A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy
GalinKa [24]

Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

Velocity component, v₁ = 5 m/s (due east)

Velocity component, v₂ = 8 m/s (due north)

Let v is the resultant of velocity. It is given by :

v=\sqrt{v_1^2+v_2^2}

v=\sqrt{5^2+8^2}=9.43\ m/s

Let E is the kinetic energy of the plane. It is given by :

E=\dfrac{1}{2}mv^2

E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2

E = 133.38 Joules

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