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sweet-ann [11.9K]
3 years ago
10

Do atoms ever touch​

Physics
2 answers:
Mumz [18]3 years ago
8 0

Answer:

nope never

Explanation:

igomit [66]3 years ago
7 0

Answer:

never ever they dont

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On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo
ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

5 0
3 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

#SPJ10

6 0
2 years ago
Convert 15 joule into erg.​
GuDViN [60]

Answer:

<h2><u>Joule</u><u>:</u></h2>

1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.

1 Joule= 1 Newton × 1 metre

1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²

So units of N is kgm/s²

So,

1 Joule

=1kgm/s² × m

=1kgm²/s²

<h2><u>Erg</u><u>:</u></h2>

1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.

1 Erg =1 Dyne × 1 cm

1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².

1 Erg=1 gmcm/s² × cm

1 Erg=1 gmcm/s² × cm=1gmcm²/s²

this is what you need to convert 1gmcm²/s² to 1kgm²/s²

<h3><u>what you need to know for conversion</u></h3>

[1gm=0.001kg

1cm²

=1cm ×1cm

=0.01 m × 0.01 m

=0.0001m²

second remains constant

]

So,

1gmcm²/s²

=0.001kg×0.0001m²/s²

=0.001kg×0.0001m²/s² =0.0000001kgm²/s²

Hence,

<h3><u>1 Erg</u><u>=</u><u>0.0000001</u><u> </u><u>Joule</u></h3><h3><u>1</u><u> </u><u>Joule</u><u>=</u><u>1</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>Erg</u></h3>

<h2>⇒15 J=15×10000000 Erg</h2><h2> =150000000 Erg</h2><h2> =1.5×10⁶ Erg</h2>
3 0
3 years ago
Read 2 more answers
Which statements about electric field lines are correct? Check all that apply.
pochemuha
They point towards a negative charge.
They point away from a positive charge.
They never cross.
3 0
3 years ago
Read 2 more answers
A 17.0 resistor and a 6.0 resistor are connected in series with a battery. The potential difference across the 6.0 resistor is m
tia_tia [17]

Answer:

V= 57.5 V

Explanation:

  • If the resistors are in the linear zone of operation, the potential difference across them, must obey Ohm's law:

        V = I*R

  • For the 6.0 Ω resistor, if the potential difference across it is 15 V, we can find the current flowing through it as follows:

       I = \frac{V}{R} = \frac{15 V}{6.0 \Omega} = 2.5 A

  • In a series circuit, the current is the same at any point of it, so the current through the battery is I = 2.5 A
  • The equivalent resistance of a series circuit is just the sum of the resistances, so, in this case, we can write the following equation:

      R_{eq} = R_{1} +R_{2} = 17.0 \Omega + 6.0 \Omega = 23.0 \Omega

  • Applying Ohm's Law to the equivalent resistance, we can find the potential difference through it, that must be equal to the potential difference across the battery, as follows:

        V = I* R_{eq}  = 2.5 A * 23.0 \Omega = 57.5 V

8 0
2 years ago
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