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3 years ago

Name a radioactive gas that is released from the ground

Physics

1 answer:

Katyanochek1 [597]3 years ago

7 0

Radon or Rn is a radioactive gas that can be released from the ground.

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Pls hurry worth 28 points <3

Kipish [7]

Answer:

Explanation:

I will GUESS that we're supposed to be looking at a plot of a position in time.

IF that is the case.

THEN the answer would be Point B because it has the steepest slope.

6 0

3 years ago

What scientific discovery did Daniel Bernoulli work on with Leonhard Euler?

Dmitriy789 [7]

He worked with Euler on elasticity and the development of the Euler-Bernoulli beam equation.

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.

7 0

3 years ago

As the distance between two charged particles increases, the force of attraction __________(increases/decreases). This is a(n) _

zubka84 [21]

Answer: decreases.

direct.

Explanation:

when you have two objects which are far apart from each other, when the distance increases the forces of attraction decreases and vice versa when the distance decreases the forces of attraction also increases.

this is a direct relationship, from careful observation and monitoring of the simulation, a pattern is noticeable by which an increase in one variable causes an opposite effect on the other variable.

4 0

3 years ago

According to the plate tectonics model what layers form earths rigid moblie plates

slamgirl [31]

The crust. The crust dude! Have you seen the layers of the earth?

3 0

3 years ago