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3 years ago

Find the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for the first 2.00 m of fall.

1 answer:

4 0

<span>Find the initial kinetic energy, the final kinetic energy, and the change in kinetic energy for the first 2.00 m of fall.</span>

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The mass of the earth is 5.96/10kg^24. The radius of the earth is approximately 6.37x10^6 calculate the force of gravity

n200080 [17]

<h2>Answer:g=9.79 $ms^{-2}$ ,A object of mass $m$ at the surface of earth experiences a force $mg$ </h2>

Explanation:

Let $M$ be the mass of earth.

Let $R$ be the radius of earth.

Let $G$ be the universal gravitational constant.

Given,

$M=5.96\times 10^{24}Kg$

$R=6.37\times 10^{6}m$

$G=6.67259 \times 10^{-11}$ $Nm^{2}Kg^{-2}$

Let $g$ be the acceleration due to gravity.

Then, $g=\dfrac{GM}{R^{2}}$

$g=\frac{6.67259 \times 10^{-11}\times 5.96\times 10^{24}}{(6.37\times 10^{6})^{2}}$

$g=9.79ms^{-2}$

A object of mass $m$ at the surface of earth experiences a force $mg$

3 0

3 years ago

According to the rule of 72 and about how many years will $78 be worth $39 if the rate of inflation is 5.8%

Monica [59]

The answer is 12.4 years

3 0

3 years ago

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A 1-kg collar (located at point (2,2) from the origin) is pulled along a vertical, frictionless bar with a force of 10 N applied

faltersainse [42]

Answer:

The acceleration of the collar is 10 m/s²

Explanation:

Given;

mass of the collar, m = 1 kg

applied force on the bar, F = 10 N

The acceleration of the collar can be calculated by applying Newton's second law of motion;

F = ma

where;

F is the applied force

m is mass of the object

a is the acceleration

a = F / m

a = 10 / 1

a = 10 m/s²

Therefore, the acceleration of the collar is 10 m/s²

3 0

2 years ago

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are char

Marta_Voda [28]

Answer:

KE = 2.535 x 10⁷ Joules

Explanation:

given,

angular speed of the fly wheel = 940 rad/s

mass of the cylinder = 630 Kg

radius = 1.35 m

KE of flywheel = ?

moment of inertia of the cylinder

$I = \dfrac{1}{2}mr^2$

= $\dfrac{1}{2}\times 630\times 1.35^2$

= 574 kg m²

kinetic energy of the fly wheel

$KE = \dfrac{1}{2}I\omega^2$

$KE =\dfrac{1}{2}\times 574\times 940^2$

KE = 2.535 x 10⁷ Joules

the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules

5 0

2 years ago

The potential energy of a catapult was completely converted into kinetic energy by releasing a small stone with a mass of 20 gra

DerKrebs [107]

1,000 grams = 1 kilogram
20 grams = 0.02 kilogram

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

(1/2) (0.02) x (15)² =

(0.01) x (225) = 2.25 joules

4 0

3 years ago

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