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2 years ago

4

Two identical metal spheres A and B are in contact. Both are initially neutral. 1.0×1012 electrons are added to sphere A, then t

he two spheres are separated. Afterward, what are the charge of A and the charge of B?

1 answer:

NISA [10]2 years ago

4 0

Answer:

$Q_A=80 nC$ , $Q_B=80 nC$ ,

Explanation:

Knowing both spheres are identical means after the equilibrium both will have equal charge so added $1.0x10^{12} e$

So to determine the charge of both spheres can use

$Q_A=Q_B$

$Q=1.6x10^{-19}C$

$Q_A=Q_B=Q_{total}$

$Q_{total}= - 1.0x10^{12}C * - 1.6x10^{-19}C$

$Q_{total}=1.6x10^{-7}C$

Both spheres are identical so

$Q_A=Q_B=\frac{Q_{total}}{2}$

$Q_A=Q_B=\frac{1.6x10^{-7}}{2}=80x10^{-9}C$

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Which of the following best describes the dependant variable?

weeeeeb [17]

Hello!

The independent variable is the variable deliberately changed.

The dependent variable is the variable that responds to change. So the answer is A.

Hope this helps. Any questions please just ask! Thank you!

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2 years ago

23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n

PSYCHO15rus [73]

Answer:

a) (95.4 i^ + 282.6 j^) N , b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

sin θ = Fjy / Fj

cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

T = 90 -32 = 58º

Fjy = Fj sin 58

Fjx = FJ cos 58

Fjx = 180 cos 58 = 95.4 N

Fjy = 180 sin 58 = 152.6 N

Andrea's force is

Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

Fx = Fjx

Fx = 95.4 N

Axis y

Fy = Fjy + Fa

Fy = 152.6 + 130

Fy = 282.6 N

F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

F² = Fx² + Fy²

F = √ (95.4² + 282.6²)

F = √ (88963)

F = 298.27 N

tan θ = Fy / Fx

θ = tan-1 (282.6 / 95.4)

θ = tan-1 (2,962)

θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

F' = 298.27 N

θ' = 180 + 71.3

θ = 251.4º

4 0

2 years ago

In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec

Bezzdna [24]

Answer with Explanation:

We are given that

$r=0.053 nm=0.053\times 10^{-9} m$

$1 nm=10^{-9} m$

Charge on proton,q= $1.6\times 10^{-19} C$

a.We have to find the electric potential of the proton at the position of the electron.

We know that the electric potential

$V=\frac{kq}{r}$

Where $k=9\times 10^9$

$V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$V=27.17 V$

B.Potential energy of electron,U= $\frac{kq_e q_p}{r}$

Where

$q_e=-1.6\times 10^{-19} c=$ Charge on electron

$q_p=q=1.6\times 10^{-19} C$ =Charge on proton

Using the formula

$U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}$

$U=-4.35\times 10^{-18} J$

8 0

2 years ago

Imagine you derive the following expression by analyzing the physics of a particular system: a=gsinθ−μkgcosθ, where g=9.80meter/

BARSIC [14]

Answer:

Explanation:

Analysis of structure gives

a=gsinθ−μkgcosθ

Notice that all the expression are right but we want to know of we can simplify the expression further.

We want to analyse if we can still further simplify the expression,

Inspecting the Right hand side of the equation, we notice that the acceleration due to gravity is common to both side, so we can bring it out i.e.

So option a is wrong because the expression can be simplified further to

a=g(sinθ−μkcosθ)

Option b is right and the best option.

Since we are given that, g=9.8m/s²

We can as well substitute that to option a

So we will have

a=9.8metre/second²(sinθ−μkcosθ)

Also option C is correct but it is not best inserting the values of g directly without simplifying the expression first

So it will have been the best option if it was written as

a=9.8metre/second²(sinθ−μkcosθ)

So the best option is B.

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2 years ago

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

a)

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

b)

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

2 years ago