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ivolga24
15 days ago
11

Rita has two small containers, one holding a liquid and one holding a gas. Rita transfers the substances to two larger container

s. Compare the behavior of the atoms in the liquid and in the gas once they are moved to the larger containers
Physics
2 answers:
Ostrovityanka [942]15 days ago
4 0

Response: Liquids conform to the shape of their container, yet they maintain a specific volume. Similarly, gases adjust their shape based on their container, but the volume of a gas is variable depending on the containment it occupies.

Explanation:

serg [1.1K]15 days ago
3 0

Response:

Liquids are able to flow easily, adapting their shape to that of the container, while retaining a constant volume. Conversely, a gas changes its form with its container, as the atoms within a gas move quickly and freely to occupy all available space. Unlike liquids, a gas's volume is adjustable based on the container it is housed in.

Explanation

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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
serg [1198]
The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
8 0
4 days ago
A bus slows down uniformly from 75.0 km/h to 0 km/h in 21 s. How far does it travel before stopping?
Yuliya22 [1153]

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

                        (10-5/12) × (21) = 218.75 meters.

7 0
15 days ago
Read 2 more answers
When a mass of 25 g is attached to a certain spring, it makes 20 complete vibrations in 4.0 s. what is the spring constant of th
kicyunya [1025]

Response: The spring constant is 25 N/m.

Details:

The body’s mass is 25 g, which converts to 0.025 kg (since 1 kg = 1000 g).

The total oscillations are 20 in 4 seconds.

Oscillations per second = \frac{20}{4}=5

Spring's frequency of vibration is = 5 s^{-1}=5 Hz

The spring constant 'k' can be derived from the relationship involving frequency, mass, and spring constant.

Frequency=\frac{1}{2\pi}\times \sqrt{\frac{k}{m}}

5 s^{-1}=\frac{1}{2\times 3.14}\times \sqrt{\frac{k}{0.025 kg}}

k=24.649 N/m\approx 25 N/m

The spring constant is 25 N/m.

3 0
7 days ago
Read 2 more answers
Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
kicyunya [1025]

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i + b j, where a and b represent constants to solve for.

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

By comparing the coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Thus,

B = 4.81452 i - 18.38908 j..... 4th quadrant

<pTherefore,

cos(Q) = 4.81452 / 12

Q = 66.346 degrees

360 - Q gives us 293.65 degrees from the + x-axis in a counterclockwise direction.

5 0
9 days ago
As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha
Softa [913]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

\Delta x\Delta p \geq \frac{\hbar}{2}

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s

\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s

6 0
6 days ago
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