Ask question

Ask question

All categories

2 years ago

5

An electron is pushed into an electric field where it acquires a 1-v electrical potential. suppose instead that two electrons ar

e pushed the same distance into the same electric field. what is the electrical potential of one of the electrons now?

2 answers:

Lapatulllka [165]2 years ago

4 0

1V electrical potential

bearhunter [10]2 years ago

4 0

The potential of the two electrons is still 1 volt, but you had to use twice as much energy to push both of them to that location.

You might be interested in

"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati

Rasek [7]

Answer:

99.95%

Explanation:

A double pulsar system named PSR J0737-3039A/B in Puppis constellation was discovered in the year 2003. Pulsars are second most densest object in the universe after black holes and they emit radio waves at regular intervals. This pair presented a great and natural setup to test the Theory of General Relativity presented by Einstein in 1915. In this theory Einstein had presented a set of equations on how the space-time fabric will be curved because of the very dense objects such as Neutron stars. It also predicted how the gravitational waves are created because of stars orbiting each other.

A team of astrophysicists led by Michael Kramer, conducted a study on how these gravitational waves will impact the time in which the radio waves emitted by pulsars will reach Earth. The result of the study proved the theory of General Relativity to be accurate up to 99.95%.

8 0

2 years ago

In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

gladu [14]

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0

2 years ago

To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes a

Gwar [14]

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

5 0

2 years ago

The weight of a bucket is 186 N. The bucket is being raised by two ropes. The free-body diagram shows the forces acting on the b

amm1812

Fnet=(115+106)-186= 34 N

mass=Force/g= 186N/9.8m/s^2 = 18.98 kg

a=fnet/mass => 34N/18.98kg = 1.79 m/s^2

so A= 1.8m/s^2

4 0

2 years ago

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

A)

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

2 years ago