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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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The initial speed of the cork was 1.57 m/s.

Explanation:

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The equation of the horizontal position of the cork in function of time is the following:

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t = time.

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If we place the origin of the frame of reference at the launching point, then x0 = 0.

We know that at t = 1.25 s, x = 1.50 m. We also know the launching angle so we can solve the equation of horizontal position for the initial speed, v0:

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What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
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Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

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