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Mazyrski [523]
2 years ago
11

An object thrown in the air has a velocity after t seconds that can be described by v(t) = -9.8t + 24 (in meters/second) and a h

eight h(t) = -4.9t 2 + 24t + 60 (in meters). The object has mass m = 2 kilograms. The kinetic energy of the object is given by K = __1 2mv2 , and the potential energy is given by U = 9.8mh. Find an expression for the total kinetic and potential energy K + U as a function of time. What does this expression tell you about the energy of the falling object?
Physics
1 answer:
marin [14]2 years ago
6 0

Answer and Explanation: Kinetic energy is related to movement: it is the energy an object possesses during the movement. it is calculated as:

K=\frac{1}{2}mv^{2}

For the object thrown in the air:

K=\frac{1}{2}.2.[v(t)]^{2}

K=(-9.8t+24)^{2}

K=96.04t^{2}-470.4t+576

Kinetic energy of the object as a function of time: K=96.04t^{2}-470.4t+576

Potential energy is the energy an object possesses due to its position in relation to other objects. It is calculated as:

U=mgh

For the object thrown in the air:

U=9.8.2.h(t)

U=9.8.2.(-4.9t^{2}+24t+60)

U=-96.04t^{2}+470.4t+1176

Potential energy as function of time: U=-96.04t^{2}+470.4t+1176

Total kinetic and potential energy, also known as mechanical energy is

TME = 96.04t^{2}-470.4t+576 + (-96.04t^{2}+470.4t+1176)

TME = 1752

The expression shows that total energy of an object thrown in the air is constant and independent of time.

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You can reason it out like this:

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-- Its average speed during that 30 sec is  (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

                                                                           =  3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

       Distance = (1/2) · (acceleration) · (time)²

                       = (1/2) ·      (8 m/s²)     · (30 sec)²

                       =      (4 m/s²)          ·      (900 sec²)

                       =            3600 meters.

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When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s².  That's about 0.82 G  !

He does zero to 60 mph in 3.4 seconds, and at the end
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He doesn't need to worry about getting a speeding ticket.
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3 0
2 years ago
The position of an object that is oscillating on an ideal spring is given by the equation x=(12.3cm)cos[(1.26s−1)t]. (a) at time
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<span>x=((12.3/100)m)cos[(1.26s^−1)t]
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 v=</span>-((12.3/100)*1.26)sin[(1.26s^−1)t]=-((12.3/100)*1.26)sin[(1.26s^−1)*(0.815)]
 v=<span> <span>-0.13261622 m/s
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6 0
2 years ago
The equilibrium fraction of lattice sites that are vacant in silver (Ag) at 600°C is 1 × 10-6. Calculate the number of vacancies
algol [13]

Answer :

The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.

Explanation:

Given,

Atomic mass of silver = 107.87 g/mol

Density of silver = 10.35 g/cm^3

Converting to g/m^3,

= 10.35 g/cm^3 × 10^6cm^3/m^3

= 10.35 × 10^6 g/m^3

Avogadro's number = 6.022 × 10^23 atoms/mol

Fraction of lattice sites that are vacant in silver = 1 × 10^-6

Nag = (Na * Da)/Aag

Where,

Nag = Total number of lattice sites in Ag

Na = Avogadro's number

Da = Density of silver

Aag = Atomic weight of silver

= (6.022 × 10^23 × (10.35 × 10^6)/107.87

= 5.778 × 10^28 atoms/m^3

The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6

= 5.778 × 10^22/m^3.

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Correct option: A

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2 years ago
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