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3 years ago

A passenger jet flies from one airport to another 1273 miles away in 2.7 h. Find its average speed.

Physics

1 answer:

AnnZ [28]3 years ago

3 0

Answer:

Average speed of passenger jet

$s = 471.48\ mi/hr$

Explanation:

Let s be the average speed of the passenger jet.

Given:

Distance from one airport to another airport $d = 1273\ mi$

Total time taken $t = 2.7\ hr$

We need to find the average speed of the passenger jet.

Solution:

Average speed formula:

The ratio of the total distance travelled by the object to the time taken to cover that distance.

$Average\ speed = \frac{Total\ distance\ travelled}{Total\ time\ taken}$

$s = \frac{d}{t}$

Substitute total distance and time value in above equation.

$s = \frac{1273}{2.7}$

$s = 471.48\ mi/hr$

Therefore, the average speed of the passenger jet is $s = 471.48\ mi/hr$ .

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3 years ago

From a cliff 37.6 m high. At the level of the sea, a rock sticks out a horizontal distance of 12.12 m. The acceleration of gravi

Aleks [24]

Answer:

4.3 m/sec

Explanation:

Here height of cliff = y = 37.6 m

Gravitational acceleration = g = 9.8 m/sec2

vi = 0 m/s

Let's find the time which the diver will take if jumps from there!

Using formula

y = vit+1/2gt2

==> 37.6= 0 + 0.5 ×9.8× $t^{2}$

==> $t^{2}$ = $\frac{37.6}{4.9}$

==> t = 2.8 sec

In this time the diver has to cover a horizontal distance of 12.12 m

If x = 12.12 m is the horizontal distance to be covered then using

x= Vx × t

==> Vx = x/t

==> Vx= 12.12/2.8 = 4.3 m/s

8 0

3 years ago

Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,

Leviafan [203]

Answer:

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

Explanation:

As we know that the charge per unit length of the long cylinder is given as

$\lambda = \frac{q}{L}$

here we know that the electric field between two cylinders is given by

$E = \frac{2k\lambda}{r}$

now we know that electric potential and electric field is related to each other as

$\Delta V = - \int E.dr$

$\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr$

$\Delta V = -2k \lambda ln(\frac{b}{a})$

$\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}$

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

7 0

2 years ago

Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$