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garri49 [273]
3 years ago
8

Having more ______________ will make a material a better conductor.

Physics
1 answer:
Tasya [4]3 years ago
6 0
I believe the answer is free electrons
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With each beat of your heart the aortic valve opens and closes. The valve opens and closes very rapidly, with a peak velocity as
Nonamiya [84]

Answer:

|Δf| = 37.3 kHz

Explanation:

given,

peak velocity = 4 m/s

speed of the sound = 1500 m/s

frequency = 7 MHz

v = C\dfrac{\pm \dlta f}{2 f_0}

\delta f = \pm 2 f_0 (\dfrac{V}{C})

\delta f = \pm 2\times 7 (\dfrac{4}{1500})

           =\pm 0.0373 MHz

           = 37.3 kHz

|Δf| = 37.3 kHz

hence, frequency shift between the opening and closing valve is 37.3 kHz

4 0
3 years ago
An increase in a sound's pitch corresponds to an increase in what other property?
Wittaler [7]
An increase in a sound's pitch corresponds to an increase in the frequency!
8 0
3 years ago
Finish the sentence
Scrat [10]
The first blank is character
The last one is eye colour.
8 0
3 years ago
There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-
andrezito [222]

Answer:

The net field will be the sum of the fields created by each charge.

where the charge Q in a position r' is given by:

E(r) = k*Q/(r - r')^2

Where k is a constant, and r is the point where we are calculating the electric field.

Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:

E₁(r) = k*3q/(r - r₁)^2

While for the other charge of -2q in the position r₂ = (d, 0, 0)

The electric field is:

E₂(r) = -k*2*q/(r - r₂)^2

Then the net field at the point r is:

E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2

E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Then if the we want to find the points r = (x, y, z) such that:

E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)

Then we must have:

0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Also remember that the distance between two points:

(x, y, z) and (x', y', z') is given by:

D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)

Then we can rewrite:

r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x + d))^2 + y^2 + z^2)

and

r - r₂ =  √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x - d))^2 + y^2 + z^2)

Replacing that in our equation we get:

0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)

0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)

We want to find the values of x, y, z such that the above equation is true.

2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)

2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]

2*(x + d)^2  + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2

2*(x + d)^2 - 3*(x - d)^2 =  3*y^2 + 3*z^2 -  2*y^2 - 2*z^2

2*(x + d)^2 - 3*(x - d)^2  = y^2 + z^2

2*x^2 + 2*2*x*d + 2*d^2 -  3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2

-x^2 + 10*x*d - d^2 = y^2 + z^2

we can rewrite this as:

- ( x^2 - 10*x*d + d^2) =  y^2 + z^2

now we can add and subtract 24*d^2 inside the parenthesis to get

- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) =  y^2 + z^2

-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2

-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2

The thing inside the parenthesis is a perfect square:

-(x - 5d)^2 + 24d^2 = y^2 + z^2

we can rewrite this as:

24d^2 = y^2 + z^2 + (x - 5d)^2

This equation gives us the points (x, y, z) such that the electric field is zero.

Where we need to replace two of these values to find the other, for example, if y = z = 0

24d^2 = (x - 5d)^2

√(24d^2)  = x - 5d

√24*d = x - 5d

√24*d + 5d = x

so in the point (√24*d + 5d, 0, 0) the net field is zero.

7 0
3 years ago
An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 43.0 km . To the
GalinKa [24]

Answer: v₂ = 5962 km

the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

Explanation:

Given that;

Lunar surface is in an altitude h = 43.0 km =  43 × 10³ m

we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²

speed of the space craft when it crashes into the lunar surface , v

decreasing speed of the space craft = 23 m/s

Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r

but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²

HERE r = Rm + h

we substitute

r = 1.74 × 10⁶ m + 43 × 10³ m

= 1.783 × 10⁶ m

On equating these, we have

G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)

v²/r = GM/r²

v = √ ( GM/r)

v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )

v = √ (2750787.9978)

v = 1658.55 m/s

Now since speed is decreasing by 23 m/s

the speed of the space craft into the lunar face is,

v₁ = 1658.55 m/s - 23 m/s

v₁ = 1635.55 m/s

Now applying conversation of energy, we say

1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)

v₂ =  √ [ v₁² + GMe (1/Rm - 1/r)]

v₂ =   √ [ 1635.55²  + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]

v₂ =  √ (2675023.8025 + 67979.24)

v₂ = √(2743003.046)

v₂ = 1656.2 m/s

now convert

v₂ = 1656.2 × 1km/1000m × 3600s/1hrs

v₂ = 5962 km

Therefore the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

8 0
3 years ago
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