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2 years ago

Please Answer Fast! Tomorrow is my exam!

Physics

1 answer:

ehidna [41]2 years ago

3 0

Average speed = Total distance/ Total time
Avg. speed = 20000/70 = 285.71
Avg. velocity = 0 as displacement is zero.

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If mechanical energy is conserved in a system the energy at any point in time can be in the form of

I am Lyosha [343]

potential, kinetic, elastc energies

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2 years ago

5. A person fishing from a pier observes that four wave crests pass by in 7.0 s and estimates that the distance between two succ

TiliK225 [7]

Answer:

v= 1.71 m/s

Explanation:

Given that

Distance between two successive crests = 4.0 m

λ = 4 m

T= 7 sec

T is the time between 3 waves.

3 waves = 7 sec

1 wave = 7 /3 sec

So t= 7/3 s

We know that frequency f

f= 1/t= 3/7 Hz

Lets take speed of the wave is v

v= f λ

f=frequency

λ=wavelength

v= 3/7 x 4 = 12 /7

v= 1.71 m/s

3 0

2 years ago

What are the positive impacts of Genetically Modified Crops?

xeze [42]

Answer:

The crops will have the ability to be resistant to certain diseases

7 0

1 year ago

Read 2 more answers

To stretch a certain nonlinear spring by an amount x requires a force F given by F = 40 x − 6 x 2 , where F is in Newtons and x

strojnjashka [21]

Answer:

64 J

Explanation:

The potential energy change of the spring ∆U = -W where W = work done by force, F.

Now W = ∫F.dx

So, ∆U = - ∫F.dx = - ∫Fdxcos180 (since the spring force and extension are in opposite directions)

∆U = - ∫-Fdx

= ∫F.dx

Since F = 40x - 6x² and x moves from x = 0 to x = 2 m, we integrate thus, ∆U = ∫₀²F.dx

= ∫₀²(40x - 6x²).dx

= ∫₀²(40xdx - 6x²dx)

= ∫₀²(40x²/2 - 6x³/3)

= ∫₀²(20x² - 2x³)

= [20x² - 2x³]₀²

= [(20(2)² - 2(2)³) - (20(0)² - 2(0)³)

= [(20(4) - 2(8)) - (0 - 0))

= [80 - 16 - 0]

= 64 J

5 0

2 years ago

In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for

4vir4ik [10]

Answer:

The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is

$k = 0.903$

Explanation:

From the question we are told that

The time constant $\tau = 3$

The potential across the capacitor can be mathematically represented as

$V = V_o (1 - e^{- \tau})$

Where $V_o$ is the voltage of the capacitor when it is fully charged

So at $\tau = 3$

$V = V_o (1 - e^{- 3})$

$V = 0.950213 V_o$

Generally energy stored in a capacitor is mathematically represented as

$E = \frac{1}{2 } * C * V ^2$

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now since capacitance is constant at $\tau = 3$

The energy stored can be evaluated at as

$V^2 = (0.950213 V_o )^2$

$V^2 = 0.903 V_o ^2$

Hence the fraction of the energy stored in an initially uncharged capacitor is

$k = 0.903$

4 0

2 years ago