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3 years ago

15. All of the following parameters are forces except

Physics

1 answer:

icang [17]3 years ago

3 0

Answer: What do you need help with?

Explanation:

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A wave has a frequency of 875 hz and a wavelength of 352 m. At what speed is this wave traveling

Sonbull [250]

Answer:

308,000 or 30.8×10^3

Explanation:

v=f×lamda

v is ?, f is 875Hz, lamda is 352m

v=875×352

v=308,000

v=30.8×10^3 m/s

5 0

3 years ago

What kind of energy does a skier have standing still at the top of a hill?

Len [333]

The skier has potential because potential energy is enery that is stored or an object that is or does not move

4 0

3 years ago

What is the largest continent

uysha [10]

Asia is the largest continent.

8 0

2 years ago

Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????

rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

$\frac{d\textup{r}}{\textup{dt}}$ = 70 cm/min

Volume of the sphere, V = $\frac{\textup{4}}{\textup{3}}\pi r^3$

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, $\frac{d\textup{V}}{\textup{dt}}$

= $\frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}$

Substituting the value of $\frac{dr}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70$

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

4 0

3 years ago

A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi

motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

3 0

3 years ago