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Alika [10]
2 years ago
11

A lorry of mass 12,000kg travelling at a velocity of 2 m/s collides with a stationary car of mass 1500kg. The vehicles move toge

ther after the impact. Calculate their velocity.
Physics
1 answer:
Lubov Fominskaja [6]2 years ago
3 0

Answer:

Approximately 1.8\; \rm m \cdot s^{-1}. Assumption: the friction between the two vehicles and the ground is negligible.

Explanation:

The momentum p of an object is the product of its mass m and its velocity v:

p = m \cdot v.

Momentum should be conserved in this collision if there's no external force on these two vehicles. In other words, the sum of the momentum of the lorry and the car should be the same before and after the collision.

Mass of the lorry: m_1 = 12000\; \rm kg.

Velocity of the lorry right before the collision: 2\; \rm m \cdot s^{-1}.

Therefore, the momentum of the lorry right before the collision would be 12000\; \rm kg \times 2\; \rm m \cdot s^{-1} = 24000\; \rm kg \cdot m\cdot s^{-1}.

Mass of the car: m_2 = 1500\; \rm kg.

Velocity of the car before the collision: 0\; \rm m \cdot s^{-1}.

Therefore, the velocity of the car before the collision would be 1500\; \rm kg \times 0\; \rm m \cdot s^{-1} = 0\; \rm kg \cdot m \cdot s^{-1}.

Sum of the momentum of the lorry and the car right before the collision:

24000\; \rm kg \cdot m \cdot s^{-1} + 0\; \rm kg \cdot m \cdot s^{-1} = 24000\; \rm kg \cdot m \cdot s^{-1}.

Under the assumption that there is no external forces on the lorry and the car, the sum of the momentum of the lorry and the car right after the collision should also be 24000\; \rm kg \cdot m \cdot s^{-1}.

Let v\; \rm m \cdot s^{-1} be the velocity of the lorry and the car after the collision. (The velocity of the two vehicles should be the same because they were moving together.)

Right after the collision, the momentum of the lorry (with a mass of 12000\; \rm kg) would be {12000\; {\rm kg} \times v\; \rm m \cdot s^{-1}} = 12000\, v \; \rm kg \cdot m \cdot s^{-1}.

Similarly, right after the collision, the momentum of the car (with a mass of 1500\; \rm kg) would be {1500\; \rm kg} \times v\; {\rm m \cdot s^{-1}} = 1500\, v\; \rm m \cdot s^{-1}.

The sum of the velocities of the two vehicles right after the collision would be:

\left(12000\, v \; {\rm kg \cdot m \cdot s^{-1}} + 1500\, v \; {\rm kg \cdot m \cdot s^{-1}}\right).

That quantity should match the momentum right before the collision. In other words:

{12000\; \rm kg \cdot m \cdot s^{-1}} = \left(12000\, v \; {\rm kg \cdot m \cdot s^{-1}} + 1500\, v \; {\rm kg \cdot m \cdot s^{-1}}\right).

Solve this equation for the velocity of the two vehicles after the collision:

\displaystyle v = \frac{12000}{12000 + 1500}\approx 1.8.

Therefore, the velocity of the two vehicles right after the collision would be approximately 1.8\; \rm m \cdot s^{-1}.

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Explanation:

Given

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North Dakota Electric Company estimates its demand trend line​ (in millions of kilowatt​ hours) to​ be: D​ = 75.0 ​+ 0.45​Q, whe
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Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

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The demand trend line​ is

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D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times101)\times0.80

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We need to calculate the demand forecast for spring

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times102)\times1.20

D=145.08\ millions\ KWH

We need to calculate the demand forecast for summer

Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

D=(75.0+0.45\times103)\times1.40

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Using given formula

D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors

Put the value into the formula

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3 0
3 years ago
PLS HELP ME OUT!!
MaRussiya [10]
In order to solve this problem, we must first find out the value of each line on the number line. However, we can make this problem more simple by ignoring every interval except for the ones between 0 and 6. There are three total intervals in between 0 and 6 (including 6 and excluding 0). Therefore, we can do 6/2, and get an interval value of 2. This means that each line adds a value of 2. Since the car is only one line past zero, we only have to add one value of 2. Since 0 + 2 = 2, our final answer is C. 2.

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Now, density is an intensive property, this means that if you have 10 grams of a given material or 1000 grams of the same material, in both cases you will find the same density.

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D = 2.5g/433.015mm^3 = 0.00577 g/mm^3

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