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2 years ago

5

a proton travelling along the x-axis is slowed by a uniform electric field E. at x = 20 cm, the proton has a speed of 3.5x10^6 m

/s and at 80cn the speed is zero. Determine the magnitude and the direction of E

1 answer:

AveGali [126]2 years ago

6 0

Answer:

The magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

Explanation:

From work-kinetic energy principles, the kinetic energy change of the proton equals the work done by the electric field.

So, ΔK = W

1/2m(v₂² - v₁²) = qEd where m = mass of proton = 1.673 × 10⁻²⁷ kg, v₁ = initial speed of proton = 3.5 × 10⁶ m/s, v₂ = final speed of proton = 0 m/s, q = proton charge = + e = 1.602 × 10⁻¹⁹ C, E = electric field and d = distance moved by proton = x₂ - x₁ , x₁ = 20 cm and x₂ = 80 cm. So, d = 80 cm - 20 cm = 60 cm = 0.6 m

1/2m(v₂² - v₁²) = qEd

E = (v₂² - v₁²)/2mqd

substituting the values of the variables into E, we have

E = ((0 m/s)² - (3.5 × 10⁶ m/s)²)/(2 × 1.673 × 10⁻²⁷ kg × 1.602 × 10⁻¹⁹ C × 0.6 m)

E = - 12.25 × 10¹² m²/s² ÷ 3.22 × 10⁻⁴⁶

E = -3.8 × 10⁵⁸ N/C

So, the magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago

A bus is moving at a speed of 150km/hr. Begins to slow at a constant rate of 3.0m/s each second. Find how far it goes before sto

ladessa [460]

Answer:

Distance = 13.9 meters

Explanation:

Given the following data;

Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s

Decelerating speed = 3m/s

To find the distance travelled with this speed;

Distance = maximum speed/decelerating speed

Distance = 41.67/3

Distance = 13.9 meters

Therefore, the bus would travel a distance of 13.9 meters before stopping.

4 0

2 years ago

A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the

Mademuasel [1]

Answer:

(a) $a=2m/sec^2$

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So $a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2$

(a) We know that force is given by

F = ma

So force will be $F=290\times 2=580N$

(b) From second equation of motion we know that

$s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m$

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

$P=\frac{W}{t}=\frac{5220}{3}=1740Watt$

(d) Time t = 1.5 sec

So $P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt$

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3 years ago

Object A has a mass of 5 kg and a velocity of 6 m/s to the east while Object 1 point B has a mass of 12 kg and velocity 0.6 m/s

Olin [163]

Answer:

Momentum of system = 37.2 Kgm/s.

Explanation:

<u>Given the following data;</u>

Mass A = 5 kg
Velocity A = 6 m/s
Mass B = 12 kg
Velocity B = 0.6 m/s

To find the momentum of the system;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

<u>For object A;</u>

Momentum A = 5 * 6

Momentum A = 30 Kgm/s

<u>For object B;</u>

Momentum B = 12 * 0.6

Momentum B = 7.2 Kgm/s

Next, we would determine the momentum of this system using the formula;

Momentum of system = Momentum A + Momentum B

Substituting the values into the formula, we have;

Momentum of system = 30 + 7.2

<em>Momentum of system = 37.2 Kgm/s.</em>

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2 years ago

Fan blades would be an analogy for which atomic model? Thomson's Bohr's electron cloud Dalton's

nata0808 [166]

The correct answer would be Electron Cloud

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2 years ago