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Svetlanka [38]
3 years ago
9

What happens when the arms of the milky move away from the center of the galaxy

Engineering
1 answer:
Alina [70]3 years ago
6 0
Well this question is though because we have never seen such a thing ! and to be quite frank when that happens , nothing good comes from it. Black Holes
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CODE
zalisa [80]

Answer:

Explanation:

I'm not 100% this is what you want, but here it is:

2

3

13

8

11

A

13

6 0
3 years ago
What is Back EMF? How does it limits the speed of a permanent magnet DC?
ss7ja [257]

Answer and Explanation:

The DC motor has coils inside it which produces magnetic field inside the coil and due to thus magnetic field an emf is induced ,this induced emf is known as back emf. The back emf always acts against the applied voltage. It is represented by E_b

The back emf of the DC motor is given by E_b=\frac{NP\Phi }{60A}

Here N is speed of the motor ,P signifies the number of  poles ,Z signifies the the total number of conductor  and A is number of parallel paths

As from the relation we can see that back emf and speed ar dependent on each other it means back emf limits the speed of DC motor

8 0
3 years ago
What is the term RF exiciter?
Sati [7]
The exciter provides fully coherent receiver local oscillator signals at radar frequency band as well as requisite, auxiliary high frequency clock signals. The exciter function is divided into an internal frequency synthesizer and an upconverter.

Hope this helps :)))
7 0
3 years ago
A manometer is used to measure the air pressure in a tanlc The fluid used has a specific gravity of 1.25, and the differentialhe
Alenkasestr [34]

Answer:

(a) 11.437 psia

(b) 13.963 psia

Explanation:

The pressure exerted by a fluid can be estimated by multiplying the density of the fluid, acceleration due to gravity and the depth of the fluid. To determine the fluid density, we have:

fluid density = specific gravity * density of water = 1.25 * 62.4 lbm/ft^3 = 78 lbm/ft^3

height = 28 in * (1 ft/12 in) = 2.33 ft

acceleration due to gravity = 32.174 ft/s^2

The change in pressure = fluid density*acceleration due to gravity*height = 78*32.174*(28/12) = 5855.668 lbm*ft/(s^2 * ft^2) = 5855.668 lbf/ft^2

The we convert from lbf/ft^2 to psi:

(5855.668/32.174)*0.00694 psi = 1.263 psi

(a) pressure = atmospheric pressure - change in pressure = 12.7 - 1.263 = 11.437 psia

(b) pressure = atmospheric pressure + change in pressure = 12.7 + 1.263 = 13.963 psia

8 0
3 years ago
A ramp from an expressway with a design speed of 30 mi/h connects with a local road, forming a T intersection. An additional lan
hram777 [196]

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

R = \dfrac{u^2}{15(e+f_s)}

where;

R= radius

f_s = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

f_s = 0.20

So;

R = \dfrac{30^2}{15(0.08+0.20)}

R = \dfrac{900}{15(0.28)}

R = \dfrac{900}{4.2}

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

5 0
3 years ago
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