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3 years ago

What is the final velocity of a 10 kg object that drops from rest and falls for 30 seconds?

1 answer:

8 0

The object is in free fall, so its velocity follows the law:
$v(t) = v_0 + gt$
where
$v_0$ is the initial velocity (zero)
g is the gravitational acceleration
t is the time

Since the initial velocity is zero, the velocity after t=30 s is
$v(30 s)=gt=(9.81 m/s^2)(30 s)=294.3 m/s$

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Answer:

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Explanation:

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3 years ago

Two frisky grasshoppers collide in midair at the top of their respective trajectories and grab onto each other, holding tight th

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Answer:

The decrease in Kinetic energy is 0.0107 Joules

Explanation:

Given

Mass of grasshoppers

Let m1 = Mass of grasshopper 1

Let m2 = Mass of grasshopper 2

Let u1 = initial speed of grasshopper 1

Let u2 = initial speed of grasshopper 2

m1 = 250g = 0.25kg

m2 = 130g = 0.13kg

u1 = 15cm/s = 0.15m/s

u2 = 65cm/s = 0.65m/s

First, we calculate the final velocity of the grasshoppers after collision using conservation of momentum.

Using

m1u1 + m2u2 = (m1 + m2) * v

Where v = final velocity

By substituton

0.25 * 0.15 + 0.13 * 0.65 = (0.25 + 0.13) * v

0.0375 + 0.0845 = 380v

0.122 = 0.38v

Make v the subject of formula

v = 0.122/0.38

v = 0.321 m/s

Calculating the Kinetic energies before and after impact.

Before collision;

KE = ½m1u1²+ ½m2u2²

KE = ½(m1u1² + m2u2²)

By substituton;

KE = ½(0.25 * 0.15² + 0.13 * 0.65²)

KE = 0.030275J

After collision:

KE = ½(m1+m2)v²

KE = ½(0.25 + 0.13) * 0.321²

KE = 0.01957779 J

Change in kinetic energy = ∆KE

∆KE = 0.030275J - 0.01957779J

∆KE = 0.01069721J

∆KE = 0.0107 J --- Approximately

Hence the decrease in Kinetic energy is 0.0107 Joules

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3 years ago

A plane travels at a speed of 205mph in still air. Flying with a tailwind, the plane is clocked over a distance of 1000 miles. F

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While plane is moving under tailwind condition it took time "t"

so here we will have

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now solving two equations

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