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3 years ago

When using a micron gauge, if the pressure gauge falls frim 1,300 to 1,000 instantly, the vehicle needs to?

1 answer:

5 0

If the micron gauge pressure falls instantly, it means that the moisture of the air-conditioning is frozen, therefore you need to move the vehicle to a warmer place in order to unfreeze the moisture.

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It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and

Trava [24]

Answer:

Explanation:

kinetic energy = 14.1 MJ = 14.1 x 10⁶ J

Let radius of flywheel be r .

volume of flywheel = π r² x t where t is thickness

= 3.14 x r² x .113 m³

= .04 r² m³

mass = volume x density

= .04 r² x 7800 = 312.73 r²kg

moment of inertia I = 1 / 2 mass x radius²

= .5 x 312.73 r² x r²

= 156.37 r⁴ kg m²

angular velocity ω = 2π x 93/60

= 9.734 rad /s

kinetic energy = 1/2 Iω² where ω is angular velocity

= .5 x 156.37 r⁴ x 9.734²

= 7408.08 r⁴

Given

7408.08 r⁴ = 14.1 x 10⁶

r⁴ = .19 x 10⁴

r = .66 x 10

= 6.60 m .

Diameter = 13.2 m

b )

centripetal acceleration of a point on its rim = ω² r

= 9.734² x 6.6

= 625.35 m /s²

5 0

3 years ago

Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small

ahrayia [7]

Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by

C = 2L = 2*pi*R ---> R = L/pi 

Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. 

we can define a small charge dq as 

dq = l*ds = l*R*da 

So the electric field can be written as: 

dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) 

dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) 

E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) 

E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)

8 0

3 years ago

One of the primary goals of the Kepler space telescope is to search for Earth-like planets. Data gathered by the telescope indic

amid [387]

Answer:

85.62 m

168.75 years

101.04 years

Explanation:

$L_0$ = Length of ship = 143 m

v = Velocity of ship = 0.8c

c = Speed of light

s = Distance to Boralis orbit = 135 ly

Gamma value

$\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \gamma=\dfrac{1}{\sqrt{1-\dfrac{0.8^2c^2}{c^2}}}\\\Rightarrow \gamma=1.67$

Length contraction is given by

$L=\dfrac{L_0}{\gamma}\\\Rightarrow L=\dfrac{143}{1.67}\\\Rightarrow L=85.62\ m$

The length is 85.62 m

Time taken

$t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{135}{0.8}\\\Rightarrow t=168.75\ years$

Time taken from the perspective one Earth is 168.75 years

Time dilation is given by

$t'=\dfrac{t}{\gamma}\\\Rightarrow t'=\dfrac{168.75}{1.67}\\\Rightarrow t'=101.04\ years$

The time taken from the perspective of the ship is 101.04 years

8 0

3 years ago

How does the density of water change when: (a) it is heated from 0o

Radda [10]

Answer:

[b] it id heated from 4o

Explanation:

3 0

3 years ago

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0

3 years ago