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Vika [28.1K]
3 years ago
13

.

Physics
1 answer:
Dmitry [639]3 years ago
5 0

Answer: 176.4 J

Explanation:

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You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
4 years ago
What shape is the unit cell of ruby?
sleet_krkn [62]
Ruby, a variety of the mineral corundum is in the trigonal crystal system, with hexagonal scalenohedra crystals
7 0
4 years ago
The greater the of an object the more force is needed to cause acceleration
d1i1m1o1n [39]

the greater the <u>mass</u> of an object the more force is needed to cause acceleration



8 0
3 years ago
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh
lara [203]

Answer:

a_y = 4.9\ m/s^2

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}

\dfrac{dz}{dy}=-0.5

The differential form with the acceleration

\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}

-0.5=-\dfrac{a_y}{a_z + g}

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

0.5=\dfrac{a_y}{0 + 9.8}

a_y = 9.8\times 0.5

a_y = 4.9\ m/s^2

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0
3 years ago
What is solar energy?
Flura [38]
I think it's D because sunlight is a solar system
3 0
3 years ago
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