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3 years ago

What happens to the moleclues within a gas when the gas Condeenses

1 answer:

7 0

Condensation happens when molecules in a gas cool down. As the molecules lose heat, they lose energy and slow down. They move closer to other gas molecules. Finally these molecules collect together to form a liquid. Hope it helps

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Zeb was lifting a box onto a moving truck. He lifted with a net force of 2000N and the box had a mass of 100 kg. What was the re

Aleksandr [31]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{2000}{100} = 20 \\$

We have the final answer as

Hope this helps you

8 0

3 years ago

Which wave has a disturbance that is parallel to the wave motion?

Leto [7]

The answer is D.<span>longitudinal</span>

6 0

3 years ago

You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver

Goryan [66]

The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
$W= \Delta U = mg \Delta h$ (1)

where m is the sister's mass, g is the gravitational acceleration and $\Delta h$ is the increase in altitude of the sister with respect to its initial position.

By calling $\theta$ the angle of the chain with respect to the vertical, the increase in altitude is given by
$\Delta h = L - L \cos \theta = L(1 - \cos \theta)$ (2)
where L is the length of the chain.

Putting (2) inside (1), we find
$W= m g L (1 - \cos \theta)$
from which we can find the mass of the sister:
$m = \frac{W}{g L (1 - \cos \theta)} = \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg$

5 0

3 years ago

Hull (1943) had rats push a lever that required 21 grams of force to budge. After they had learned to push the lever in order to

Zina [86]

Years of research have demonstrated that rats are intelligent creatures who experience pain and pleasure, care about one another, are able to read the emotions of others, and would assist other rats, even at their own expense.

<h3>Experiments:</h3>

In trials carried out at Brown University in the 1950s, rats were trained to press a lever for food, but they stopped pressing the lever when they noticed that with each press, a rat in an adjacent cage would scream in pain (after experiencing an electric shock).

Rats were trained to press a lever to lower a block that was hanging from a hoist by electric shocks administered by experimenters. A rat was subsequently hoisted into a harness by the experimenters, and according to their notes, "This animal normally shrieked and wriggled sufficiently while dangling, and if it did not, it was jabbed with a sharp pencil until it exhibited indications of discomfort." Even if it wasn't in danger of receiving a shock, a rat watching the scenario from the floor would pull a lever to lower the hapless rodent to safety.

Learn more about experiments on rats here:

brainly.com/question/13625715

#SPJ4

6 0

2 years ago

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

$m_1*a+m_1*g=m_2*g-m_2*a$

Solving a, we have

$(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2$

We can then replace this value of a in one for the sums of force and find the tension T:

$T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N$

5 0

3 years ago