When do phase transitions occur in molecules?

The velocity of a ball changes from ‹ 9, −6, 0 › m/s to ‹ 8.96, −6.12, 0 › m/s in 0.02 s, due to the gravitational attraction of

Alenkasestr [34]

Answer:

a) $a=(-2,-6,0)m/s^2$ , with a magnitude of $6.3m/s^2$

b) $\frac{\Delta p}{\Delta t}=(-0.24,-0.72,0)Kgm/s^2$ , with a magnitude of $0.76Kgm/s^2$

c) $F=(-0.24,-0.72,0)N$ , with a magnitude of $0.76N$

Explanation:

We have:

$v_{ix}=9m/s, v_{iy}=-6m/s, v_{iz}=0m/s\\v_{fx}=8.96m/s, v_{fy}=-6.12m/s, v_{fz}=0m/s\\t=0.02s, m=0.12Kg$

We can calculate each component of the acceleration using its definition $a=\frac{\Delta v}{\Delta t}$

$a_x=\frac{v_{fx}-v_{ix}}{t} = \frac{(8.96m/s)-(9m/s)}{0.02s} =-2m/s^2\\a_y=\frac{v_{fy}-v_{iy}}{t} = \frac{(-6.12m/s)-(-6m/s)}{0.02s} =-6m/s^2\\a_y=\frac{v_{fz}-v_{iz}}{t} = \frac{(0m/s)-(0m/s)}{0.02s} =0m/s^2\\$

The rate of change of momentum of the ball is $\frac{\Delta p}{\Delta t} = \frac{\Delta mv}{\Delta t} = \frac{m\Delta v}{\Delta t} = ma$

So for each coordinate:

$\frac{\Delta p_x}{\Delta t}=-0.24Kgm/s^2\\\frac{\Delta p_y}{\Delta t}=-0.72Kgm/s^2\\\frac{\Delta p_z}{\Delta t}=0Kgm/s^2$

And these are equal to the components of the net force since F=ma.

If magnitudes is what is asked:

$a=\sqrt{a_x+a_y+a_z} =6.3m/s^2\\F=ma=\frac{\Delta p}{\Delta t}=0.76N$

3 0

3 years ago

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi

valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

$v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s$

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

$x=vt=(15.1m/s)(13.49s)=203.69m$ (1)

Next you calculate the distance that teacher has traveled for 13.6s:

$x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m$ (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

$x_{min}=692.88m-203.69m=489.19m$

5 0

3 years ago

Which statement BEST describes the condition of steady-state equilibrium? - System slowly adjusts to long-term changes in input

krok68 [10]

Answer:

System inputs and outputs fluctuate around a stable average so the system does not move far from its average condition.

Explanation:

Steady-state equilibrium can also be called dynamic equilibrium. The main difference between the two is the type of system we are considering. When a system is closed, and it attains equilibrium, there is no transfer of energy. In the case of an open system, even if the system achieves equilibrium there will be some transfer of energy but it will not deviate far from it's equilibrium point, that is, it will be in a steady-state.

6 0

3 years ago

If the volleyball hits the net and comes over to land on the other side it is?

photoshop1234 [79]

Other teams point ......

8 0

3 years ago

Read 2 more answers

A girl with a mass of 27 kg is playing on a swing. There are three main forces

N76 [4]

The tension in the swing's chain at the bottom of the swing is 178.35 N.

The given parameters:

Mass of the girl, m = 27 kg
Speed of the girl, v = 3 m/s
Radius of the circle, r = 4 m

The tension in the swing's chain at the bottom of the swing is calculated as follows;

$T = mg + ma_c\\\\ T= mg + \frac{mv^2}{r} \\\\T = (12 \times 9.8) + (\frac{27 \times 3^2}{4} )\\\\T = 117.6 \ N \ + \ 60.75 \ N\\\\T = 178.35 \ N$

Thus, the tension in the swing's chain at the bottom of the swing is 178.35 N.

Learn more about tension in vertical circle here: brainly.com/question/19904705

3 0

2 years ago