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crimeas [40]
3 years ago
15

A compressed spring in a toy is used to launch

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

Kinetic energy = (1/2) (mass) x (speed²)

If the ball was launched in the horizontal direction,
then its total kinetic energy was

                               (1/2) x (0.005 kg) x (5 m/s)²

                           =  (0.0025 x 25)  (kg-m²/sec²)

                           =     0.0625 Joule .

And that would be the amount of energy stored in the spring.

If the ball was launched at some angle above the horizontal, then
it had some vertical speed too.  Its total kinetic energy was then
some greater number, and there was more energy stored in
the spring.
 
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Answer:

Speed

Explanation:

The answer is speed. It doesn’t affect potential energy because it’s associated with position. But since kinetic energy is associated with motion speed affects it.

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Find A and effective resistance.<br> (Ill give Brainliest if you provide explaination)
Alex777 [14]

Answer:

A = 2.4 A

R_{eq}  = 5 \ \Omega

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through  R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

R_{series \, 1} = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

R_{series \, 2} = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

\dfrac{1}{R_{eq}}  = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}

Therefore;

R_{eq} =  \dfrac{R_{series \, 1} \times  R_{series \, 2}}{R_{series \, 1} + R_{series \, 2}}

Therefore;

R_{eq} =  \dfrac{10\times  10}{10 + 10} \ \Omega = 5 \ \Omega

R_{eq}  = 5 \ \Omega

The value of the current, <em>A</em>, in the circuit, I = V/R_{eq}

A = I = 12 V/(5 Ω) = 2.4 A

A = 2.4 A

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The qualifications boil down to: College education.

In most university or industrial research organizations, you might be able to work there as a member of the team who doesn't get much pay or much respect, with research going on all around you directed by other people, after you've gotten you Master's degree.

But you really don't have a shot at leading anything, or having much to say about what's being researched or how, until you have a PhD degree in the field where you'd like to do the research.

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