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Usimov [2.4K]
3 years ago
6

Need some help with these plz

Engineering
1 answer:
pashok25 [27]3 years ago
5 0

100: D, third law of motion

101: D, second law of motion

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When does the vc-turbo engine use lower compression ratios?.
Veronika [31]

Answer:

Explanation:

The VC-T engine (for "variable compression, turbocharged") can adjust its compression ratio between 8:1 and 14:1 on the fly, offering high-compression efficiency under light loads and the low compression needed for turbocharged power under hard acceleration.

7 0
2 years ago
Who was part of dempwolf his firm when he first started
Zinaida [17]

Explanation:

Dempwolf created by John Augustus, Among the most prominent innovative solutions in Southern California Pennsylvania was established by Dempwolf with  brother Reinhardt or uncle's son Frederick entered the company of J.A. Dozens of structures in 10 states were engineered by Dempwolf.

5 0
2 years ago
When a variable is stored in memory, it is associated with an address. To obtain the address of a variable, the & operator c
liubo4ka [24]

Answer:

Explanation:

1) C program file addressOfScalar.c

#include <stdio.h>

int main()

{

//intialize a char variable, print its address and the next address

char charvar = 'a';

printf("address of charvar = %p\n", (void *)(&charvar));

printf("address of charvar - 1 = %p\n", (void *)(&charvar - 1));

printf("address of charvar + 1 = %p\n", (void *)(&charvar + 1));

//intialize a int variable, print its address and the next address

int intvar = 1;

printf("address of intvar = %p\n", (void *)(&intvar));

printf("address of intvar - 1 = %p\n", (void *)(&intvar - 1));

printf("address of intvar + 1 = %p\n", (void *)(&intvar + 1));

}

In C programming language, an int variable takes 4 bytes of memory. So any arithmetic on integer address, always considers it as 4 bytes of data. So intvar-1 refers to a location 4 bytes before intvar's address and intvar+1 refers to 4 bytes after intvar's address.

3 0
3 years ago
Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

 Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

  h=3374 KJ/Kg  ,s=6.59 KJ/Kg-K  (Super heated steam table)

<u>At 10 KPa:</u>

h_g=2675.1 KJ/Kg, h_f=417.51  KJ/Kg

s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= h_f+x(h_g- h_f)

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34  KJ/Kg

Lets take m is the mass flow rate of steam

So 5\times 10^3=m\times (3374-2449.34)

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0
3 years ago
public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fracti
True [87]

Answer:

The Full details of the answer is attached.

7 0
3 years ago
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