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spin [16.1K]
2 years ago
8

Newtons second law lab report link

Physics
1 answer:
gulaghasi [49]2 years ago
5 0

Answer:

ghittu iihg उह्स उउह्स उग्य्किव जिक्ह्ब

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To avoid an accident, a driver steps on the brakes to stop a 1000kg car traveling at 65km/h. If the braking distance is 35m, how
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To stop the car it would be 100m because if the car is going to 65km/h then it would still be 100km/h
8 0
3 years ago
A rock is thrown from a bridge at an angle 30∘ below horizontal.immediately after the rock is released, is the magnitude of its
Wittaler [7]
<span>The magnitude of the rock is equal to g. After the rock is released, there are no more forces acting on it, yet gravity remains. The initial inputs, on a bridge, at an angle of 30 deg below horizontal do not matter after the release.</span>
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3 years ago
If a displacement vector has a negative x component and a positive y component,
bezimeni [28]

Answer:

The direction will be NE i.e North east.

Explanation:

From the question, it will be travelling in North east direction because the the negative X components in vector is in horizontal direction which is the Eastern direction when using the four cardinal points and the Y components, the positive is in North direction if you use the four cardinal points. Therefore, the vector with both X components and Y components will be travelling in Northeast direction.

8 0
3 years ago
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

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