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2 years 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

1 answer:

Lena [83]2 years 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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A spaceship flies from Earth to a distant star at a constant speed. Upon arrival, a clock on board the spaceship shows a total e

m_a_m_a [10]

Answer:

35 288 mile/sec

Explanation:

This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:

$10 years = \frac{2d}{v}$

If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:

$0.8 = \sqrt{1-\frac{v^{2} }{c^{2} } }$

Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:

$0.8 = \sqrt{1 - \frac{v^{2} }{c^{2} } }$

$0.64 = 1-\frac{v^{2} }{186282^{2} }$

solving for v, gives = 35 288 miles/s

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2 years ago

A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet

kogti [31]

The centripetal force, Fc, is calculated through the equation,
Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius.
Substituting the known values,
Fc = (112 kg)(8.9 m/s)² / (15.5 m)
= 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N.

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2 years ago

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A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

$v(t)=At+Bt^2$

Differentiating with respect to time

$\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt$

At t = 0

$1.4=A$

Hence, A = 1.4 m/s²

$B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3$

B = -0.10493 m/s³

At t = 5 seconds

$a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2$

a = 1.29507 m/s²

$T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N$

T = 28095.8271 N

Weight of rocket

$W=2530\times 9.81=24819.9\ N$

$\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W$

T = 1.13198 W

3 0

2 years ago

Problem: An ice hockey player hits a puck of mass 0.15 kilograms with a force of 100 newtons in the horizontal direction. What i

valina [46]

The acceleration produced in a body is always in the direction of the resultant force acting on the body. Therefore, we may determine the horizontal acceleration using the horizontal force applied. To do this, we may apply the mathematical form of Newton's second law:

Force = mass * acceleration
acceleration = force / mass

Substituting the values,
a = 100 / 0.15
a = 666.7 m/s²

The acceleration of the hockey puck is 670 m/s²

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2 years ago

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If a rainstorm drops 5 cm of rain over an area of 13 km2 in the period of 3 hours, what is the momentum (in kg · m/s) of the rai

stiks02 [169]

To solve the problem it is necessary to apply the concepts related to Conservation of linear Moment.

The expression that defines the linear momentum is expressed as

P=mv

Where,

m=mass

v= velocity

According to our data we have to

v=10m/s

d=0.05m

$A=13*10^6m^2$

Volume $(V) = A*d = (15*10^6)(0.03) = 3.9*10^5m^3$

t = 3hours=10800s

$\rho = 1000kg/m^3$

From the given data we can calculate the volume of rain for 5 seconds

$V' = \frac{V}{t}*\Delta t_{total}$

Where,

$\Delta t_{total}$ It is the period of time we want to calculate total rainfall, that is

$V' = \frac{3.9*10^5}{10800}*5$

$V' = 1.805*10^2m^3$

Through water density we can now calculate the mass that fell during the 5 seconds:

$m' = V'*\rho$

$m' = 1.805*10^2*1000$

$m' = 1.805*10^5m^2$

Now applying the prevailing equation given we have to

$P=m'v$

$P = (1.805*10^5)(10)$

$P = 1.805*10^6 Kg.m/s$

Therefore the momentum of the rain that falls in five seconds is $1.805*10^6 Kg.m/s$

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2 years ago