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Misha Larkins [42]
3 years ago
7

While driving his 600-kg ultra-compact electric car, a distracted driver starts crossing a drawbridge. Not realizing that the dr

awbridge is about to open, he ends up in the river below with his car. Fortunately the car stayed afloat for a while and he is able to escape before it starts sinking. (a) If the volume of the interior is 2.0 m3, what fraction of the car is immersed while it floats? (b) Eventually, water starts to leak in and gradually fills the car. What is the mass of the water that entered the car when the car starts to sinks? Density of water is 1000 kg/m3.
Physics
1 answer:
stellarik [79]3 years ago
4 0

Answer:

Explanation:

a )

weight of car = 600 kg .

According to Archimedes principle , for free floatation ,

weight of car = weight of displaced water

600 x 9.8 = v x 10³ x 9.8 , v is volume of water displaced and 10³ kg/ m³ is density of water.

v = .6 m³

required fraction

= .6 / 2

= .3

b )

Let mass of water entering before the start of sinking of car .

mass of maximum displaced volume of water by car = 2 x 10³m³.

(600 + m ) x g = 2 x 10³ x g         [   10³ is density of water    ]

600 + m = 2000

m = 1400 kg .  

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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
3 years ago
A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric fiel
Free_Kalibri [48]

Answer:

B=2.74\times 10^{-10}\ T

Explanation:

It is given that,

A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m

We need to find the magnetic vector of the wave at the point P at that instant.

The relation between electric field and magnetic field is given by :

c=\dfrac{E}{B}

c is speed of light

B is magnetic field

B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T

So, the magnetic vector at point P at that instant is 2.74\times 10^{-10}\ T.

3 0
3 years ago
Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh
ahrayia [7]

Answer:

i) FALSE,  ii) TRUE,  iii) FALSE, iv)  FALSE

Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

           n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

Let's apply these equations to our case where

          nₓ <n_y

i) The expression of the refractive index

          nₓ < n_y

         \frac{c}{v_x} = \frac{c}{n_y}

         v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

 the statement of FALSE

8 0
2 years ago
On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel
Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}

|v_{0.2}|=11.99 m/s\approx 12 m/s

Velocity of 0.4 kg mass

v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}

|v_{0.4}|=12.99 m/s\approx 13 m/s

Thus total Kinetic energy =\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}

Kinetic energy=48.2 J

8 0
3 years ago
An object is 30 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine the location of the
poizon [28]

Answer:

Inverted

Real

Explanation:

u = Object distance =  30 cm

v = Image distance

f = Focal length = 10 cm

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{30}\\\Rightarrow \frac{1}{v}=\frac{1}{15}\\\Rightarrow v=15\ cm

As, the image distance is positive the image is real and forms on the other side of the lens

m=-\frac{v}{u}\\\Rightarrow m=-\frac{-15}{30}\\\Rightarrow m=-0.5

As, the magnification is negative the image is inverted

3 0
2 years ago
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