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3 years ago

State the relationship between degrees Celsius and kelvins

1 answer:

3 0

Kelvin is a unit of absolute temperature and is often used in measurements and it can also be associated with degrees Celsius. 0 Kelvin is equivalent to -273.15 degrees Celsius.

In order to solve for either unit of temperature, the equation is:

T(Kelvin)= T(degrees Celsius) + 273.15
T(degrees Celsius) = T(Kelvin) - 273.15

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A student is using a metal wire to toast marshmallows in a campfire. Which statement best explains why the wire feels hot after

Licemer1 [7]

Answer:

Heat is transferred from the fire through the wire by conduction.

Explanation:

8 0

3 years ago

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

klio [65]

Answer:

Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

$100\; \rm J$ of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Kinetic energy before the collision:

First ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

Mass of the two balls, combined: $0.5\; \rm kg$ .
Velocity of the two balls right after the collision: $60\; \rm m\cdot s^{-1}$ .

Sum of the kinetic energies of the two balls right after the collision:

$\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J$ .

Therefore, $1000\; \rm J - 900\; \rm J = 100\; \rm J$ of kinetic energy would be lost during this collision.

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4 years ago

University Physics 9.68: Accelerating Compact Disc. A computer disc drive is turned on starting from rest and has constant angul

faust18 [17]

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4 years ago

Someone on Earth would measure the star Vega to be 25 light years away. If you fly a spaceship to Vega at 0.999c, you will measu

gladu [14]

If you fly a spaceship to Vega at 0.999c, you will measure the distance to be much less than 25 light years.

Most space objects use light-years to represent their distance. Light-years are the distance that light travels during the year on Earth. Light-years are about 6 trillion miles (9 trillion km). This is a 6 followed by 12 zeros.

At the same speed, a movement equivalent to one light-year takes about 11.3 billion days. Life expectancy for Americans is currently estimated at 78.74 years, which is equivalent to 28,740 days. Therefore, to get there, you need to live about 400,000 times as long as the average American.

Learn more about light years here: brainly.com/question/1224192

#SPJ4

4 0

2 years ago

The gravitational force between Pluto and Charon is 3.61 × 1018 N. Pluto has a mass of 1.3 × 1022 kg, which is only slightly gre

raketka [301]

<u>Answer:</u>

<em>The distance between Pluto and Charon is $1.96 \times 10^7 m$ </em>