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3 years ago

Which of the following has the most thermal energy?

1 answer:

8 0

Answer:
2. A 1 litre mug of hot chocolate at 75 degrees.

Explanation:
Thermal energy is directly proportional to mass, so as the mass increases, the thermal energy of the substance increases as well.

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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

Lubov Fominskaja [6]

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 $1.8\; \rm m \cdot s^{-1}$ .

3 0

2 years ago

Leap years _____. happen because the Earth revolves around the sun in less than 365 days make up for the extra one-fourth day th

anzhelika [568]

The Earth takes (365 and 1/4) days to orbit the sun. That's the length of our 'year'.

Since it's not really possible to make our calendar a quarter of a day longer every year, we make the calendar a <u><em>whole</em></u> day longer every 4 years.

So "Leap years" make up for the extra one-fourth day the Earth takes to orbit the sun.

7 0

3 years ago

For a given ideal gas if it temperature is constant the pressure of the gas is proportional ??

svetlana [45]

Answer:

C) Inversely to its volume

Explanation:

As the volume decreases (Container gets smaller) the pressure increases. Conversely, If the volume increases (Container gets larger) the pressure decreases.

5 0

2 years ago

Read 2 more answers

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm