Some____

Two players find themselves in a legal battle over a patent. The patent is worth 20 for each player, so the winner would receive

Alborosie

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

5 0

3 years ago

Multiple Choice

ra1l [238]

I need more details to your question

4 0

3 years ago

Read 2 more answers

What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?

jeka57 [31]

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

6 0

3 years ago

A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee

Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist = 42 s

solution

we know initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at

so acceleration a will be

a = $\frac{25-0}{8}$

a = 3.125 m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 × t

S2 = 25 × 16

S2 = 400 m = 0.4 km

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = $\frac{S}{t}$

velocity of motorist = $\frac{500}{42}$

velocity of motorist = 42.857 km/h

3 0

3 years ago

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

3 years ago

Some_____