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vazorg [7]
3 years ago
15

The direction of an electric field is the direction (5 points) a negative test charge will move in the field a positive test cha

rge will move in the field either a positive or negative test charge will move in the field the test charge may be either negative or positive
Physics
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

a positive test charge will move in the field

Explanation:

The direction of an electric field corresponds to the direction of motion of a positive test charge in the electric field. In fact:

- the electric field produced by a positive charge points outwards the charge --> this is because when a positive test charge is placed in this field, it will feel a repulsive force (because two positive charges repel each other), so it will move away from the positive charge that produces the field

- the electric field produced by a negative charge points towards the charge  itself--> this is because when a positive test charge is placed in this field, it will feel an attractive force (because a positive and a negative charge attract each other), so it will move toward the negative charge that produces the field.

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In 1977 off the coast of Australia, the fastest speed by a vessel on the water
fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration a_{ave} is the variation of velocity  \Delta V over a specified period of time  \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

a_{ave}=1.80 m/s^{2}

\Delta V=V_{f}-V_{o} being V_{o}=0 the initial velocity and V_{f} the final velocity

\Delta t=85.6 s

Then:

a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}

Since V_{o}=0:

a_{ave}=\frac{V_{f}}{\Delta t}}

Finding V_{f}:

V_{f}=a_{ave} \Delta t

V_{f}=(1.80 m/s^{2})(85.6 s)

Finally:

V_{f}=154.08 m/s

8 0
3 years ago
Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt
Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

= \frac{4}{3}\pi(6400+ 50)^3 -  \frac{4}{3}\pi (6400)
^3\\\\=  \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3

Hence the volume of atmosphere is 2.6\times 10^{19} m^3

(b)

Write the ideal gas equation as foll ows:

PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles

no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules


Hence the required molecules is 13.3\times10^{43} molecules


(c)

Write the ideal gas equation as follows:

PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles

no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules

Hence the required molecules in Caesar breath is 1.2\times 10^{23} molecules

(d)

Volume fraction in Caesar last breath is as follows:  

Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}

(e)

Since the volume capacity of the human body is 500 mL.

Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath

5 0
3 years ago
In a tug of war, side A applies 10N force and side B applies 8N .which side will the rope move
choli [55]
The side B will fall down so that means the rope will move the direction of A
7 0
3 years ago
If you double the voltage in a circuit and reduce the resistance by a factor of four, what will happen to the current?
Gennadij [26K]

Answer:

It will increase by a factor of 8.

Explanation:

By Ohm's Law, we can relate current, voltage and resistance. It is expressed as V=IR. That is, there is a direct relationship between voltage and resistance and voltage and current.

V = IR

V1/2V1 = I1R1 / I2R1/4

1/2 = 4I1/I2

I2 = 8I1

3 0
3 years ago
A 0.091-in-diameter electrical wire at 90°F is covered by 0.02-in-thick plastic insulation (k = 0.075 Btu/h·ft·°F). The wire is
olya-2409 [2.1K]

Answer:

The critical radius of the plastic insulation is 0.72 inches.

Explanation:

Given that,

Diameter = 0.091 in

Thickness = 0.02 in

Initial temperature = 90°F

Final temperature = 50°F

Heat transfer coefficient = 2.5 Btu/h.ft²°F

Material conductivity = 0.075 Btu/h.ft °F

We need to calculate the critical radius of the plastic insulation

Using formula of critical radius

r_{cr}=\dfrac{2K}{h}

Where, k = Material conductivity

h = Heat transfer coefficient

Put the value into the formula

r_{cr}=\dfrac{2\times0.075}{2.5}

r_{cr}=0.06\ ft

r_{cr}=0.72\ inches

Hence, The critical radius of the plastic insulation is 0.72 inches.

4 0
3 years ago
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