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den301095 [7]
3 years ago
12

A kangaroo can jump over an object 2.50 m high. (a) Considering just its vertical motion, calculate its vertical speed when it l

eaves the ground. (b) How long a time is it in the air?
Physics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

<h2>a) Vertical speed when it leaves the ground is 7 m/s</h2><h2>b) Kangaroo is in air for 1.43 seconds.</h2>

Explanation:

a) We have equation of motion v² = u² + 2as

Initial velocity, u = ?

Acceleration, a = -9.81 m/s²  

Final velocity, v = 0 m/s  

Displacement, s = 2.50 m

Substituting  

v² = u² + 2as

0² = u² + 2 x -9.81 x 2.50

u = 7 m /s

Vertical speed when it leaves the ground is 7 m/s

b)  We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 7 m/s

        Acceleration, a = -9.81 m/s²  

        Displacement, s = 0 m   --- Time is calculating for returning to ground    

     Substituting

                      s = ut + 0.5 at²

                      0 = 7 x t + 0.5 x -9.81 x t²

                      t = 1.43 seconds

Kangaroo is in air for 1.43 seconds.

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I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!
sergij07 [2.7K]

Answer:

<h3>a)</h3>

\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}

  • V = 12 V
  • P = 24 W

\implies \mathsf{24=\frac{12^2}{R} }

\implies \mathsf{24R=12^2 }

\implies \mathsf{24R=144 }

<u>=> R= 6 Ohms(Ω)</u>

<h3>b)</h3>

\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}

  • Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

  • V = 240 V

\implies \mathsf{100=\frac{240^2}{R} }

<em>R and 100 can interchange places</em>

\implies \mathsf{R=\frac{240^2}{100} }

\implies \mathsf{R=\frac{57600}{100} }

<u>=> R = 576 Ω</u>

<u></u>

By Ohm's Law:

\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}

=> 240 = I × 576

=>

=> I = 0.417 A

<h3 /><h3>c)</h3>

I don't know it's resistance,... so sorry

<h3>d)</h3>

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

\boxed{\mathfrak{Power \propto  \frac{1}{Resistance} }}

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

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