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liq [111]
3 years ago
5

Define one standard meter, one standard kilogram and one second

Physics
1 answer:
lidiya [134]3 years ago
6 0
<h3>The metre is the length of the path travelled by light in vacuum during a time interval of 1299 792 458 of a second.</h3>

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Plz help<br>I need it fast<br>​
Rufina [12.5K]

Answer:

perpendicular to

Explanation:

it means perpendicular to .....should u come across something like this / / , this one means parallel to .....

7 0
3 years ago
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If ram is 40kg and travels 200m in 30 sec find his power​
WARRIOR [948]

Work done

  • N×200
  • mg200
  • 40(10)(200)
  • 400(200)
  • 80000J

Power

  • Work done/Time
  • 80000/30
  • 8000/3
  • 2668W
8 0
2 years ago
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Are forces of attraction among the particles in all mater? true or false.
Luden [163]

Answer:

the answer is true

Explanation:

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2 years ago
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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
A weather balloon is filled to a volume of 6000 L while it is on the ground, at a pressure of 1 atm and a temperature of 273 K.
avanturin [10]

Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L

Solution :

Using combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 1 atm

P_2 = final pressure of gas = 0.3 atm

V_1 = initial volume of gas = 6000 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 273 K

T_2 = final temperature of gas = 240 K

Now put all the given values in the above equation, we get the final pressure of gas.

\frac{1atm\times 6000L}{273K}=\frac{0.3atm\times V_2}{240K}

V_2=17582.4L

Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L

4 0
3 years ago
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