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zhenek [66]
3 years ago
13

Which states of matter have particles that move independently of one another with very little attraction

Physics
1 answer:
Crazy boy [7]3 years ago
6 0
The state of matter that the particles move independently of one another with very little attraction is, I believe, gas
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A golf club exerts an average horizontal force of 1000 n on a 0.045 -kg golf ball that is initially at rest on the tee. the club
OlgaM077 [116]
The impulse (the variation of momentum of the ball) is related to the force applied by
\Delta p = F \Delta t
where \Delta p is the variation of momentum, F is the intensity of the force and \Delta t is the time of application of the force. 
Using F=1000 N and \Delta t = 1.8 ms=1.8 \cdot 10^{-3}s, we can find the variation of momentum:
\Delta p = (1000 N)(1.8 \cdot 10^{-3} s)=1.8 kg m/s

This \Delta p can be rewritten as
\Delta p = p_f - p_i = mv_f - mv_i
where p_f and p_i are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
\Delta p = mv_f
from which we find the final velocity of the ball:
v_f =  \frac{\Delta p }{m}= \frac{1.8 kg m/s}{0.045 kg}=  40 m/s
8 0
3 years ago
As an astronaut goes out into space, her mass _______________ and her weight ____________________
mel-nik [20]
Her mass stays the same and her weight changes
6 0
3 years ago
Read 2 more answers
9. A wave on Beaver Dam Lake passes by two docks that are 40.0 m apart.
Kobotan [32]

Answers:

a) 10 m

b) time=1.6 s, frquency=0.625 Hz

c) 6.25 m/s

Explanation:

a) If there is a crest at each dock and another three crests between the two docks, and the wavelength \lambda is the distance between to crests; this means we have 4\lambda in 40 m:

40 m=4\lambda

Clearing \lambda:

\lambda=\frac{40 m}{4}

\lambda=10 m

b) This part can be solved by a Rule of Three:

If 10 waves ---- 16 s

1 wave ----- T

Then:

T=\frac{(1 wave)(16 s)}{10 waves}

T=1.6 s This is the period of the wave

On the other hand, the frequency f of the wave has an inverse relation with its period T:

f=\frac{1}{T}

f=\frac{1}{1.6 s}

f=0.625 Hz This is the frequency of the wave

c) The speed v of a wave is given by the following equation:

v=\frac{\lambda}{T}

v=\frac{10 m}{1.6 s}

Finally:

v=6.25 m/s

4 0
3 years ago
what would the net force be on the box in the problems shown below.( both force and direction).​ for all four diagrams. please e
DIA [1.3K]

Answer:
A) object moves 20 N [West] or -20 N [East]
B) object moves 6 N [South] or -6 N [North]
C) object moves 90 N [West] or -90 N [East]
D) object does not move and is at rest*

*Rest means 0


Why:

A)both forces from north and south that are pushing against the object neutralize each other. Assume that north is positive and south is negative: 20 [N] + (-20) [S] = 0
On West and east, you can see that west has a greater force. Assume that west is negative and east is positive: 50 [E] + (-70) [W] = -20 [E]
8 0
3 years ago
A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hit
lapo4ka [179]

Answer:

The  angle is  \theta  =  15.48^o

Explanation:

From the question we are told that  

     The distance of the dartboard from the dart is  d  =  3.66  \ m

     The time taken is  t =  0.455 \ s

   

The  horizontal component of the speed of the dart is mathematically represented as

      u_x =  ucos \theta

where u is the the velocity at dart is lunched

  so

      distance =  velocity \ in \ the\  x-direction  *  time

substituting values

      3.66 =   ucos  \theta *  (0.455)

 =>   ucos \theta =  8.04  \ m/s

From projectile kinematics the time taken by the dart can be mathematically represented as

         t  =  \frac{2usin \theta }{g}

=>    usin \theta =  \frac{g  * t}{2 }

       usin \theta =  \frac{9.8  * 0.455}{2 }

      usin \theta = 2.23

=>   tan \theta =  \frac{usin\theta }{ucos \theta }  =  \frac{2.23}{8.04}

       \theta  =  tan^{-1} [0.277]

      \theta  =  15.48^o

     

4 0
3 years ago
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