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meriva
2 years ago
14

I need help

Physics
2 answers:
kupik [55]2 years ago
7 0
I like to just keep writing them over and over on a page also if you remember them just before the exam than as soon as you start write them on the front of your test so you don’t forget them.
Hope this helps.
Fantom [35]2 years ago
3 0
Try and make mnemonics out of them. So like h is happy (hydrogen) and so on?
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WILL MARK BRAINLIEST!!
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air above a wild fire

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A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass
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Answer:

Explanation:

a ) It is given that bomb was at rest initially , so ,  its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis  after explosion of third part   = mv cosθ

= 10 v cosθ

Momentum along x of first part = -  5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

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Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 +  10 v sinθ

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squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

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2 years ago
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Two parallel wires are separated by 6.10 cm, each carrying 2.85 A of current in the same direction. (a) What is the magnitude of
Westkost [7]

Answer:

The force per unit length is 2.66 \times 10^{-5} \ N/m

Explanation:

The current carrying by each wires = 2.85 A

The current in both wires flows in same direction.

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Now we will use the below expression for the force per unit length. Moreover, before using the below formula we have to change the unit centimetre into meter. So, we just divide the centimetre with 100.

F/l = \frac{\mu _0i_1 i_2}{2\pi d} \\i_1 = 2.85 \\i_ 2 = 2.85  \\\mu _0 = 4\pi \times 10^{-7} \\d = 0.061 \\F/l = \frac{4\pi \times 10^{-7} \times 2.85 \times 2.85}{2 \pi \times 0.061} \\= 2.66 \times 10^{-5} \ N/m

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