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Ber [7]
1 year ago
8

A charge Q is distributed uniformly along the x axis from x1 to x2. What would be the magnitude of the electric field at x0 on t

he x-axis? Assume that ke = 1 4 π ǫ0 and x0 > x2 > x1 for a Coulomb constant of 8.98755 × 109 N · m2 /C 2
Physics
1 answer:
Lena [83]1 year ago
8 0

Answer:

  E =  k Q    1 / (x₀-x₂) (x₀-x₁)

Explanation:

The electric field is given by

              dE = k dq / r²

In this case as we have a continuous load distribution we can use the concept of linear density

              λ= Q / x = dq / dx

              dq = λ dx

We substitute in the equation

           ∫ dE = k ∫ λ dx / x²

We integrate

           E = k λ (-1 / x)

We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁

           E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)

           E = k λ  (x₂ -x₁) / (x₀-x₂) (x₀-x₁)

We replace the density

             E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]

             E =  k Q    1 / (x₀-x₂) (x₀-x₁)

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If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus
anastassius [24]

Answer:

E=477.92\ W.m^{-2}

Explanation:

Given that:

Absolute temperature of the body, T=273+30=303\ K

  • emissivity of the body, \epsilon=1

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

E=\epsilon. \sigma.T^4

where:

\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}   (Stefan Boltzmann constant)

Now putting the respective values:

E=1\times 5.67\times 10^{-8}\times 303^4

E=477.92\ W.m^{-2}

5 0
1 year ago
If gravity between the Sun and Earth suddenly vanished, Earth would continue moving in
Ksenya-84 [330]

Answer:

Earth would continue moving by uniform motion, with constant velocity, in a straight line

Explanation:

The question can be answered by using Newton's first law of motion, also known as law of inertia, which states that:

"an object keeps its state of rest or of uniform motion in a straight line unless acted upon by an external net force different from zero"

This means that if there are no forces acting on an object, the object stays at rest (if it was not moving previously) or it continues moving with same velocity (if it was already moving) in a straight line.

In this problem, the Earth is initially moving around the Sun, with a certain tangential velocity v. When the Sun disappears, the force of gravity that was keeping the Earth in circular motion disappears too: therefore, there are no more forces acting on the Earth, and so by the 1st law of Newton, the Earth will continue moving with same velocity v in a straight line.

6 0
1 year ago
A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n
Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The<u> trajectory</u> is <u>the path followed by the body, the distance it travelled</u> (is a scalar quantity).  

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).  

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m).  Being its displacement 95 m.

8 0
1 year ago
If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti
Ann [662]
We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km
8 0
2 years ago
A spaceship of frontal area 10 m2 moves through a large dust cloud with a speed of 1 x 106 m/s. The mass density of the dust is
Step2247 [10]

Answer:

The decelerating force is 3\times 10^{- 11}\ N

Solution:

As per the question:

Frontal Area, A = 10\ m^{2}

Speed of the spaceship, v = 1\times 10^{6}\ m/s

Mass density of dust, \rho_{d} = 3\times 10^{- 18}\ kg/m^{3}

Now, to calculate the average decelerating force exerted by the particle:

Mass,\ m = \rho_{d}V                                (1)

Volume, V = A\times v\times t

Thus substituting the value of volume, V in eqn (1):

m = \rho_{d}(Avt)

where

A = Area

v = velocity

t = time

m = \rho_{d}(A\times v\times t)                  (2)

Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t

From Newton's second law of motion:

F = \frac{dp}{dt}

Thus differentiating w.r.t time 't':

F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}

where

F_{avg} = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N

4 0
1 year ago
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