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nexus9112 [7]
4 years ago
11

an object experiences buoyant force of 15.9 n when immersed in a fluid of density 917 kg/m^3. what is the volume of the object?

Physics
1 answer:
77julia77 [94]4 years ago
5 0

Answer:

volume = 1.02m³

Explanation:

F= density x g

g = F/density

g = 15.9/917 = 0.017m/s²

F = m x g

15.9 = m x 0.017

m = 15.9/0.017 = 935.3kg

density = mass/volume

917 = 935.3/volume

volume = 935.3/917 = 1.02m³

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It is called the Kuiper Belt.
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4 years ago
The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of
goldfiish [28.3K]

Answer:

6.6*10^{27}e/m^3

Explanation:

When we are dealing with Hall voltage, it is necessary to have the values of the current, the magnetic field, the length, the area and the number of carriers at hand. The Hall voltage equation is given by,V_h = \frac{iB}{neL}

Where,

i= current

B= Magnetic field

L = Length

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We need replace and solve for n,

n= \frac{iB}{V_h e L}

n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}

n= 6.6*10^{27}electron.m^{-3}

Therefore the density of charge carrier is 6.6*10^{27}e/m^3

5 0
4 years ago
a motorcycle traveling on the highway at a velocity of 120 kilometre-per-hour passes a car traveling at a velocity of 90 kilomet
musickatia [10]

Answer:

The relative velocity of the motorcycle to a passenger in the car is 30 km/h

Explanation:

The question relates to the principle of relative velocity and reference frames

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The velocity of the car, v₂ = 90 km/h

The relative velocity of an object X with regards to another object Y is the velocity the object X will seem to be moving with to an observer in the rest frame of object Y written as \underset{v}{\rightarrow}_{X|Y} = \underset{v}{\rightarrow}_{X} - \underset{v}{\rightarrow}_{Y}

Therefore, the relative velocity of the motorcycle to the car is v_{1|2} = v₁ -  v₂, which give;

v_{1|2}  = 120 km/h - 90 km/h = 30 km/h

The relative velocity of the motorcycle to a passenger in the car = 30 km/h.

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3 years ago
An antibaryon composed of two antiup quarks
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Answer:

(2) −1 e

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The antiparticle of up quark is antiup quark and has charge -\frac{2}{3}e charge.

The antiparticle of down quark is antidown quark and has charge +\frac{1}{3}e charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e

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