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3 years ago

A peacock is flying around and its velocity

Physics

1 answer:

rodikova [14]3 years ago

4 0

Answer:

2 m.

Explanation:

From the question given, the following data were obtained:

Initial velocity (u) = 2 m/s

Initial time = 2 s

Final velocity (v) = 4 m/s

Final time (t) = 3 s

Next, we shall determine the change in velocity. This can be obtained as follow:

Initial velocity (u) = 2 m/s

Final velocity (v) = 4 m/s

Change in velocity (Δv) =?

Change in velocity (Δv) = v – u

Change in velocity (Δv) = 4 – 2

Change in velocity (Δv) = 2 m/s

Next, we shall determine the change in time. This can be obtained as follow:

Initial time (t1) = 2 s

Final time (t2) = 3 s

Change in time (Δt) =..?

Change in time (Δt) = t2 – t1

Change in time (Δt) = 3 – 2

Change in time (Δt) = 1 s

Finally, we shall determine the displacement (Δx) of the peacock as follow:

Velocity (Δv) = 2 m/s

Time ((Δt) = 1 s

Displacement (Δx) =?

Velocity = Displacement /Time

ΔV = Δx/Δt

2 = Δx /1

2 = Δx

Δx = 2 m

Therefore, the displacement of the peacock is 2 m

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1 year ago

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77julia77 [94]

<span>choices are:
all male citizens could own property
an established caste system
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ans are:
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frutty [35]

Answer:

Explained

Explanation:

When the barbell is accelerated upward, the force exerted by the athlete is greater than the weight of the barbell. (The barbell, simultaneously, pushes with greater force against the athlete.) When acceleration is downward, the force supplied by the athlete is less than the force applied by the barbell on the athlete.

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3 years ago

What is the period of oscillation of a mass 40kg on a spring with constant k=10N/m?

iVinArrow [24]

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Explanation:

The period of oscillation of a spring-mass system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

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k is the spring constant

For the system in this problem, we have

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Substituting into the equation, we find

$T=2\pi \sqrt{\frac{40}{10}}=12.56 s$

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