All categories

3 years ago

How can any student outside apply for studying engineering at Cambridge University

1 answer:

3 0

Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!

You might be interested in

A 3-phase , 1MVA, 13.8kV/4160V, 60 Hz, transformer with Y-Delta winding connection is supplying a3-phase, 0.75 p.u. load on the

Tanya [424]

Answer:

a) 23.89 < -25.84 Ω

b) 31.38 < 25.84 A

c) 0.9323 leading

Explanation:

A) Calculate the load Impedance

current on load side = 0.75 p.u

power factor angle = 25.84

$I_{load}$ = 0.75 < 25.84°

attached below is the remaining part of the solution

B) Find the input current on the primary side in real units

load current in primary = 31.38 < 25.84 A

C) find the input power factor

power factor = 0.9323 leading

attached below is the detailed solution

8 0

3 years ago

A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the

sergeinik [125]

Answer:

1123.6 pounds/ square inch.

Explanation:

Fluid pressure is the ratio of force or weight applied by the fluid per unit area.

i.e Fluid pressure = $\frac{weight}{area}$

The maximum load of the jack is obtained at its maximum capacity = 5000 pounds

Area of the large piston on the jack = 4.45 square inches

Thus,

Fluid pressure = $\frac{5000}{4.45}$

= 1123.5955

Fluid pressure = 1123.6 pounds/ square inch

Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.

7 0

3 years ago

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th

Sergeeva-Olga [200]

Answer:

HEAT LOST

polycarbonate = 252 W

soda lime glass = 1680 W

aerogel = 16.8 W

COST associated with heat loss

polycarbonate = $ 262.08

soda lime glass = $ 1,747.2

aerogel = $ 17.472

The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

Explanation:

Given that;

surface area for each window = 0.4m * 0.4m = 0.16m^2

DeltaT = 90°C, L = 12mm = 0.012m

thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER

so at 300K

KsL = 1.4 W/mK

Kag = 0.014 W/mK

Kpc = 0.21 W/mK

Now HEAT LOSS

for polycarbonate;

Qpc = -KA dt/dx

NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)

so Qag = (0.21 * 0.16 * 90) / 0.012

= 252 W

for soda lime glass;

Qsl = (1.4 * 0.16 * 90) / 0.012

= 1680 W

for aerogel

Qaq = (0.014 * 0.16 * 90) / 0.012

= 16.8 W

Now for COST associated with heat lost

for polycarbonate;

cost = Qpc * 130 * 8 * 1/1000

= 252 * 130 * 8 * 1/1000

= $ 262.08

for soda lime glass;

cost = 1680 * 130 * 8 * 1/1000

= $ 1,747.2

for aerogel

cost = 16.8 * 130 * 8 * 1/1000

= $ 17.472

Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

6 0

3 years ago

I need the solution

Alinara [238K]

Answer:

Isn't the answer written upside down in the sentence?

Explanation:

7 0

2 years ago

The water in a soil flows from Point K to Point L, a distance of 250 ft. Point K is at elevation 543 ft and Point L is at elevat

Over [174]

Answer:

0.124

Explanation:

We calculate the hydraulic gradient by the formulas below.

I = (change in h)/(change in l)-----eqn 1

I = (hk-hl)/change in L ----- equation 2

At k the headloss = hk,

At L the headloss = hL

The distance of water travel is change in I

Total head at k

hk = 543+23

= 566 ft

Total head at L

hL = 461+74

= 535 ft

Change in L = 250

When we substitute these values in equation 2

566-535/250

= 0.124

The hydraulic gradient is 0.124

3 0

3 years ago

How can any student outside apply for studying engineering at Cambridge University​

How can any student outside apply for studying engineering at Cambridge University