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2 years ago

Two cars, one of mass 1400 kg, and the

Physics

1 answer:

Vikki [24]2 years ago

3 0

Suppose that, in the x-y plane, the first car is moving to the right so that its velocity is given by the vector

v₁ = (14 m/s) i

and the second car is moving upward so that its velocity vector is

v₂ = (20 m/s) j

Then the total momentum of two cars before their collision is

m₁v₁ + m₂v₂ = (1400 kg) (14 m/s) i + (2300 kg) (20 m/s) j

= (19,600 i + 46,000 j) kg•m/s

Their momentum after the collision is

(1400 kg + 2300 kg) v = (3700 kg) v

where v is the velocity vector of the wreckage.

By conservation of momentum,

(19,600 i + 46,000 j) kg•m/s = (3700 kg) v

Let a and b be the horizontal and vertical components of v, respectively. Then

19,600 kg•m/s = (3700 kg) a ⇒ a ≈ 5.2973 m/s ≈ 5.3 m/s

46,000 kg•m/s = (3700 kg) b ⇒ b ≈ 12.4324 m/s ≈ 12 m/s

so that the final speed of the wreckage is

||v|| = √(a² + b²) ≈ 13.5139 m/s ≈ 14 m/s

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3 years ago

Create a model in the space below that demonstrates how action potentials ensue. A sample model can be found under the Unit Pack

Pepsi [2]

Answer:

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Explanation:

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2 years ago

An object is located 51 millimeter from a diverging lens. te object has a hight of 13 millimeters and the image height is 3.5 mi

Sati [7]

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3 years ago

A stone is thrown vertically upward with a speed of 12m/s from the edge of a cliff 70 m high (a) How much later it reaches the b

jonny [76]

Answer

given,

vertical speed of stone,v = 12 m/s

height of the cliff = 70 m

a) time taken by the stone to reach at the bottom of the cliff

We know that,

S = u t + 1/2 a t²

- 70 = 12 t - 0.5 x 9.8 t²

4.9 t² - 12 t - 70 = 0

solving the equation

t = 5.2 s (neglecting the negative value)

b) again using equation of motion

v = u + a t

v = 12 - 9.8 x 5.2

v = -38.96 m/s

ignoring the negative sign

magnitude of velocity is equal to 38.96 m/s

c) total distance travel by the stone

vertical distance covered by the stone

v² = u² + 2 g h

0 = 12² - 2 x 9.8 x h

h = 7.34 m

to reach the stone to the same level distance travel be doubled.

Total distance travel by the stone

H = h + h + 70

H = 7.34 x 2 + 70

H = 84.7 m.

8 0

2 years ago

car 1 is traveling south at 18 m/s and has a full load, giving it a total mass of 14,650 kg. Car 2 is traveling north at 11 m/s

kvasek [131]

Answer:

v_{4}= 80.92[m/s] (Heading south)

Explanation:

In order to calculate this problem, we must use the linear moment conservation principle, which tells us that the linear moment is conserved before and after the collision. In this way, we can propose an equation for the solution of the unknown.

ΣPbefore = ΣPafter

where:

P = linear momentum [kg*m/s]

Let's take the southward movement as negative and the northward movement as positive.

$-(m_{1}*v_{1})+(m_{2}*v_{2})=-(m_{1}*v_{3})+(m_{2}*v_{4})$

where:

m₁ = mass of car 1 = 14650 [kg]

v₁ = velocity of car 1 = 18 [m/s]

m₂ = mass of car 2 = 3825 [kg]

v₂ = velocity of car 2 = 11 [m/s]

v₃ = velocity of car 1 after the collison = 6 [m/s]

v₄ = velocity of car 2 after the collision [m/s]

$-(14650*18)+(3825*11)=(14650*6)-(3825*v_{4})\\v_{4}=80.92[m/s]$

4 0

2 years ago