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Aneli [31]
3 years ago
14

What happens to the attraction between molecules as the temperature increases or decreases?

Physics
1 answer:
Damm [24]3 years ago
7 0
The inter molecular force of attraction between molecules decreases with increase in temperature. Because due to increase in temperature, the kinetic energy of the molecules increase and they starts moving with higher velocity.
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The headlights of a car are 1.6 m apart and produce light of wavelength 575 nm in vacuum. The pupil of the eye of the observer h
Xelga [282]

To solve this problem it is necessary to apply the concepts related to angular resolution, for which it is necessary that the angle is

\theta = 1.22\frac{\lambda}{nd}

Where

d = Diameter of the eye

n = Index of refraction

D = Distance between head lights

\lambda= Wavelength

Replacing with our values we have that

\theta = 1.22 \frac{(1.22)(575*10{-9})}{1.4(4*10^{-3})}

\theta = 1.252*10^{-4}rad

Using the proportion of the arc length we have to

L = \frac{D}{\theta}

Where L is the maximum distance, therefore

L = \frac{1.6}{1.252*10^{-4}}

L = 12.77km

Therefore the maximum distance from the observer that the two headlights can be distinguished is 12.77km

4 0
3 years ago
The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.
iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0
3 years ago
Read 2 more answers
Can someone please help me out?
maria [59]

Explanation:

If you want to get speed, u have to divided distance over time

The lowest speed will lose

8 0
4 years ago
Read 2 more answers
A car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver n
satela [25.4K]

Answer:

a) t1 = v0/a0

b) t2 = v0/a0

c) v0^2/a0

Explanation:

A)

How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0

Vf = 0

Vf = v0 - a0*t

0 = v0 - a0*t

a0*t = v0

t1 = v0/a0

B)

How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.

at this point

U = 0

v0 = u + a0*t

v0 = 0 + a0*t

v0 = a0*t

t2 = v0/a0

C)

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.

t1 = t2 = t

Distance covered by the train = v0 (2t) = 2v0t

and we know t = v0/a0

so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0

now distance covered by car before coming to full stop

Vf2 = v0^2- 2a0s1

2a0s1 = v0^2

s1 = v0^2 / 2a0

After the full stop;

V0^2 = 2a0s2

s2 = v0^2/2a0

Snet = 2v0^2 /2a0 = v0^2/a0

Now the separation between train and car

= (2v0^2)/a0 - v0^2/a0

= v0^2/a0

8 0
4 years ago
In a mass spectrometer, a single-charged particle (charge e) has a speed of 1.0 × 10 6 m/s and enters a uniform magnetic field o
Nonamiya [84]

Answer:

The mass is  m  =6.4*10^{-28} \ kg

Explanation:

From the question we are told that

   The  speed of the charge is  v   = 1.0 *10^{6} \  m/s

    The  magnetic field is  B = 0.20 \ T

     The radius is r  =  0.02 \ m

      The value of the charge is  e  = 1.60 *10^{-19} \  C

The centripetal acting on the charge moving in the circular orbit is mathematically represented as

        F_c  =  \frac{mv^2}{r }

Now this centripetal force is due to the force exerted on the charge by the magnetic field on the charge which is mathematically represented as

     F_m  =  qv  B  sin\theta

At the maximum of this magnetic force \theta =  90 ^o

So  

     F_m  =  e v  B  sin(90)

      F_m  =  e v  B

Now given that it is this  magnetic force that is causing the circular motion we have that

       F_c  =  F_m

=>     \frac{mv^2}{r }  =  ev  B

=>     m  = \frac{e * B  *  r  }{v }

substituting values

       m  = \frac{ 1.60 *10^{-19} *  0.20   *  0.020   }{1.0*10^{6} }

     m  =6.4*10^{-28} \ kg

8 0
4 years ago
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