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kogti [31]
4 years ago
13

Can a supertanker float in a bathtub?

Physics
2 answers:
zhuklara [117]4 years ago
6 0
1. Has to be a very large bathtub or very, very small supertanker 2. The car hit harder because the bug is not going anywhere near the same speed as the car, therefore it is not exerting the same force. 3. It's so fast it's hard to put a number on it and and it depends on the think Es of wire, distance from wall, and which type of T.V. (because if it requires less electricity then the electrons will travel slower) 4. The ball being thrown, because when and object is thrown and then falls, it always falls at a greater speed then those just being dropped.
Ugo [173]4 years ago
5 0
I'm so glad you asked !

1).  If the bathtub is big enough to accommodate the length, width, and draft
of the ship, and the tub is full of water, then the ship floats in the tub.
(Strange as it may seem, the same can be said of the planet Saturn.)

2).  Momentum is conserved.  ===> The total momentum of (car + bug)
is the same before and after the collision.  ===> The impulse imparted
to the car is equal to the impulse imparted to the bug.  ===> F₁·t = F₂·t .
===> For every action, there is an equal and opposite reaction.
===> The force exerted on the bug is equal to the force exerted on the car.
===> The bug sustains more damage than the car does.

3).  This simple-sounding question is actually a very complicated question.
... You said "an electron".  Do you mean the same electron ?  Or do you mean
the electric current ?
... The 'drift velocity' of the same electron is only millimeters per hour.
... If you're plugged into a common AC wall socket, then a single electron is
pulled this way, then that way, 60 times a second, and never gets anywhere.
... Electric current flows by means of one electron bumping the next one,
which bumps the next one, which bumps the next one, which bumps the
next one, and so on all along the wire.  So the EFFECT shows up at the
other end of the wire much faster than any single electron gets there.
... That speed depends on the thickness and composition of the wire,
the thickness and composition of the insulation around the wire, the
distance between the two wires in the power cord, and some other things.
It can range anywhere from 45% to 99% of the speed of light.

-- If your power cord is 5 feet long and the propagation velocity is 0.99c,
then the answer to your question is  0.000 000 005 13 second.

-- If your power cord is 10 feet long and the propagation velocity is 0.45c,
then the answer to your question is   0.000 000 022 6 second.

The actual situation is somewhere in that range.

4).  When you toss a ball or a stone straight up, it goes up for a while
as it slows down, then turns around, then falls for a while as it speeds up. 
It's fairly easy to prove that it returns to your hand with the same speed
as you tossed it.
So if you toss two stones, with the same speed, from the same height,
one up and one down, then the one that went up and came back has the
same speed as it passes your hand going down. 
They both hit the ground with the same speed.
But there IS one difference:  The one you tossed UP hits the ground LATER !
You might be interested in
HELP ASAP PLEASE SCIENCE
shutvik [7]

Answer:

I think it's claim 1 because the same mass would cause them both to either fall off or to stay there

3 0
3 years ago
Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co
mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copperV=8.0\times10^{-6}\ m^3

Volume of liquid V=7.6\times10^{-6}\ m^3

Latent heat of fusion L= 13.05 kJ

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}

Put the value into the formula

\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}

\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}

\Delta T=4.162\ K

Hence, The change in the equilibrium melting point is 4.162 K.

5 0
3 years ago
Anya recorded the temperatures of four different smooth materials after they were placed under a heat lamp for thirty minutes. A
bekas [8.4K]

Answer:

B

Explanation:

4 0
4 years ago
Read 2 more answers
At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t
kari74 [83]

Answer:

The road bank angle is 16.38⁰.

Explanation:

radius of curvature of the road, r = 50 m

allowable speed of car on the road, v = 12 m/s

The bank angle is calculated as;

\theta = tan^{-1} (\frac{v^2}{gr} )

where;

θ is the road bank angle

g is acceleration due to gravity = 9.8 m/s²

\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0

Therefore, the road bank angle is 16.38⁰.

8 0
3 years ago
A proton moves through an electric potential created by a number of source charges. Its speed is 2.5X105 m/s at a point where th
yanalaym [24]

Answer:

(6.67 × 10⁵) m/s

Explanation:

What will be the proton's speed a short time later when it reaches a point where the potential is -500V.

According to the law of comservation of energy, the sum of potential and kinetic energy of the proton at any point in time is always constant.

P.E₁ + K.E₁ = P.E₂ + K.E₂

P.E₁ = U₁ = qV₁

q = charge on the proton = (1.602 × 10⁻¹⁹) C

V₁ = 1500 V

P.E₁ = U₁ = (1.602 × 10⁻¹⁹) × (1500) = (2.403 × 10⁻¹⁶) J

K.E₁ = (1/2)mv₁²

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v₁ = (2.5 × 10⁵) m/s

K.E₁ = (1/2)(1.673×10⁻²⁷)(2.5 × 10⁵)² = (5.23 × 10⁻¹⁷) J

P.E₂ = U₂ = qV₂

q = charge on the proton = (1.602 × 10⁻¹⁹) C

V₂ = -500 V

P.E₂ = U₂ = (1.602 × 10⁻¹⁹) × (-500) = (-8.01 × 10⁻¹⁷) J

K.E₂ = ?

P.E₁ + K.E₁ = P.E₂ + K.E₂

(2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) = (-8.01 × 10⁻¹⁷) + K.E₂

K.E₂ = (2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) + (8.01 × 10⁻¹⁷) = (3.727 × 10⁻¹⁶) J

K.E₂ = (1/2)mv₂²

m = (1.673 × 10⁻²⁷) kg

v₂ = ?

(3.727 × 10⁻¹⁶) = (1/2)(1.673 × 10⁻²⁷)v₂²

v₂² = (4.455 × 10¹¹)

v₂ = 667,457.863839 m/s = (6.67 × 10⁵) m/s

Hope this Helps!!!!

6 0
3 years ago
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