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ser-zykov [4K]
3 years ago
14

Supposing you were in space in a weightless environment, would it require a force to set an object in motion

Physics
1 answer:
Jobisdone [24]3 years ago
7 0

Yes.  If your smartphone was floating in front of your face, motionless
relative to you, it would require a force to start it moving toward you or
away from you. 

But there's no minimum force required.  ANY force, no matter how small,
even smaller than the smallest force that you can imagine, would set it in
motion. 

The thing is, though, that the smaller the force acting on it, the smaller
acceleration it would get, and the slower it would move away from where
it is. 

So if, say, you wanted to send it across the crew compartment and over
to your sleeping bag on the wall, and you had all day to watch it mope
along over there, you might breathe on it, and the force of your breath
would set it in slow motion in that direction.  But if you wanted to throw it
at your crewmate, you'd need to give it more force.
  
You might be interested in
Help please i will mark brainlist!!
Dmitrij [34]
The answer is A.


p=m/v
p= 240/60
p= 4 g/cm^3
7 0
3 years ago
Read 2 more answers
A sinusoidal wave is traveling along a rope. The oscillator that generates the wave completes 40.0 vibrations in 30.0s . A given
Phantasy [73]

From the solution that I have done, the wavelength in the question that we have is 31.88 cm

<h3>How to solve for the wavelength</h3>

The frequency in the question is given as 40/30 = 1.33 hz

Next we have to solve for V

= 425/10

= 42.5 cm/s

v = frequency * wavelength

we have to put in the values in the formula. This would be

42.5 = 1.33 x wavelength

we have to divide through by 1.33 to get the wavelength. This would be

42.5/1.333 = wavelength

31.88 cm = wavelength

Hence we can say that the wavelength in the question that we have here is 31.88 cm

Read more on wavelength here:

brainly.com/question/10728818

#SPJ4

8 0
1 year ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.1 m/s. How long does h
balandron [24]
We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)

Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)

-0.31=11.1t+(1/2)(-9.81)t^2
Rearrange and solve for t
-4.905t^2+11.1t-0.31=0
t=-0.02756 or t=2.291 seconds 
Reject the negative root to give
t=2.29 seconds (to 3 significant figures)

3 0
3 years ago
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

7 0
4 years ago
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