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2 years ago

Two cars collide and come to a complete stop. where did all of their energy go?

1 answer:

3 0

Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.

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A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass

STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is

   y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is   y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

      = - (4 π² M / 15²) sin(2π t/15)

      = - 0.1755 M sin(2π t/15) .

There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

0.08333 G = 0.1755 M

The 'M' is what we need to find.

Divide each side by 0.1755 : M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
   M = (0.08333 / 0.1755) (9.8) = 4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0

2 years ago

If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th

devlian [24]

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

7 0

2 years ago

our iclicker operates at a frequency of approximately 900 mhz (900x106 hz). what is the approximate wavelength of the em wave pr

Elena-2011 [213]

wavelength of the EM wave produced by your iclicker is 0.33 m.

<h3>What makes an EM wave?</h3>

When an electric field (illustrated in red arrows) combines with a magnetic field, electromagnetic waves are generated (which is shown in blue arrows). An electromagnetic wave's magnetic and electric fields are perpendicular to each other and to the wave's direction.
A changing magnetic field causes a changing electric field, and vice versa—the two are inextricably related. Electromagnetic waves are created by changing fields. Electromagnetic waves, unlike mechanical waves, do not require a medium to propagate.

The clicker emits EM (electromagnetic) wave which travels at the speed of light, that is

v = 3 x 10⁸ m/s

The frequency is

f = 900mHz = 9 x 10⁸ Hz

velocity = frequency * wavelength, the wavelength, λ, is given by

fλ = v

λ = v/f

= (3 x 10⁸ m/s) / (9 x 10⁸ 1/s)

= 1/3 m = 0.333 m

To learn more about electromagnetic waves refer,

brainly.com/question/25847009

#SPJ1

6 0

9 months ago

The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.55 g of conden

SSSSS [86.1K]

Answer:

m = 62.14 g

Explanation:

Energy used to melt the ice is the energy released by the condensation of the water forms on the glass

so here we have

energy for the condensation of water is given as

let mass of water condensed = m

$E = m_1 L_f$

now the energy of vaporization is given as

$E = m_2 L_v$

here we know that

$L_f = 79.8 kCal/kg$

$L_v = 580 k Cal/kg$

Now we have

$8.55 \times 580 = m \times 79.8$

$m = 62.14 g$

3 0

3 years ago

A capacitor consists of two metal surfaces separated by an insulating layer. A new capacitor has no charge on either of its surf

mihalych1998 [28]

<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>

When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a measure of the ability of a capacitor to store energy. </em>For any capacitor, the capacitance is a constant defined as:

$C=\frac{Q}{V_{ab}}$

To maintain $C$ constant, if Q increases V also increases.

On the other hand, the potential energy $U$ can be expressed as:

$U=\frac{Q^{2}}{2C}$

In conclusion, as Q increases the potential energy also increases.

5 0

2 years ago