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2 years ago

At what separation will two charges, each of magnitude 6.0fYC,exert a force of 1.4N on each other?

1 answer:

5 0

I assume that "fYC" is just a writing mistake and the two charges have magnitude $q=6 \mu C=6 \cdot 10^{-6}C$ .
The electrostatic force between the two charges is
$F=k_e \frac{q^2}{d^2}$
where $k_e = 8.99 \cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant, q is the magnitude of the two charges, and d is the separation between them.

We know the value of the force, F=0.14 N, so re-arranging the formula and using these data we can solve to find the value of d:
$d= \sqrt{ \frac{k_e q^2}{F} } =0.48 m$

So, the separation between the two charges is 0.48 m.

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James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

Over [174]

Answer:

$4.1\cdot 10^8 N$

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

$g=9.8 m/s^2$ is the gravitational acceleration

$h=11,000 m$ is the depth

Substituting,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

And so, the inward force exerted on it is

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

2 years ago

Read 2 more answers

A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto

daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 = $741.68/yr

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0

2 years ago

three neutral metal cans mounted on isulating stands are touching a negatively charge ballon is brought near can a can b is then

Nimfa-mama [501]

Charge on can A is positive.
Charge on can C is negative.
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon.
As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.

7 0

2 years ago

gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an

Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0

2 years ago

A 6.0-μF capacitor charged to 50 V and a 4.0-μF capacitor charged to 34 V are connected to each other, with the two positive pla

ch4aika [34]

Answer:

5702.88 J or 5.7mJ

Explanation:

Given that :

C 1 = 6.0-μF

C 2 = 4.0-μF

V 1 = 50V

V 2 = 34V

Note that : Q = CV

Q 1 = C1 * V1

Q 1 = 50×6 = 300μC

Q 2 = 34×4 = 136μC

Parallel connection = C 1 + C 2

= 6+4 = 10μC

V = Qt/C

Where Qt = Q1+Q2

V = Q1+Q2/C

V = 300+136/10

V = 437/10

V = 43.6volts

Uc1 = 1/2×C1V^2

= 1/2 × 6μF × 43.6^2

= 1/2 × 6μF × 1900.96

= 3μF × 1900.96volts

= 5702.88J

= 5702.88J/1000

= 5.7mJ

4 0

2 years ago