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3 years ago

Everyone why are you reporting my answers i didnt do anything to you

Engineering

2 answers:

MrRa [10]3 years ago

8 0

ANSWER;Did You Post anything bad?

masya89 [10]3 years ago

5 0

Answer:

IDK

Explanation:

same thing is happening to me

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a metal coin has certain properties that can be measured.which property of a coin is different on the moon that is on earth?

Sloan [31]

Answer:

Coins weigh less on the Moon.

Explanation:

Gravity is only 1/6th as strong on the Moon than it is on Earth. Where a nickle is about 5 grams on Earth, it is less than 1 gram on the Moon. Gravity is affected by the size of the planet or moon. The Moon is much less massive than the Earth.

8 0

3 years ago

PLEASE ANSWER ASAP!

Butoxors [25]

Answer:

Please check the explanation.

Explanation:

While driving, white lines would require us to stay within the lane, also marking the shoulder of the roadway.

While yellow lines generally mark the center of a two-way road being used for two-way traffic.

We generally can pass on a two-way road if the yellow centerline is broken.

7 0

3 years ago

Read 2 more answers

The acceleration due to gravity on the surface of the moon is 1.62m/s^2. The moon's radius is Rm+1738km. A) What is the weight i

Anastaziya [24]

Answer:

weight is 12.6 N

force is 4.05 N

Explanation:

given data

acceleration = 1.62 m/s²

radius = 1738 km

mass = 10 kg

distance = 1738 km

to find out

weight and force

solution

we apply here weight formula that is

weight = mass × acceleration ...................1

put here value

weight = 10 × 1.26

weight = 12.6 N

and

force = mass × An

force = 10 × 1.62 (1738/ 1738+1738)² = 4.05 N

so force is 4.05 N

4 0

3 years ago

Saturated liquid-vapor mixture of water, called wet steam, in a steam line at 1500 kPa is throttled to 50 kPa and 100°C. What is

frozen [14]

Answer:

$x = 0.944$

Explanation:

Steam at outlet is an superheated steam, since $T > T_{sat}$ . From steam tables, the specific enthalpy is:

$h_{out}=2682.4\,\frac{kJ}{kg}$

The throttle valve is modelled after the First Law of Thermodynamics:

$h_{in} = h_{out}$

Hence, specific enthalpy at inlet is:

$h_{in}=2682.4\,\frac{kJ}{kg}$

The quality in the steam line is:

$x = \frac{2682.4\,\frac{kJ}{kg}-844.55\,\frac{kJ}{kg}}{2791.0\,\frac{kJ}{kg} - 844.55\,\frac{kJ}{kg} }$

$x = 0.944$

6 0

3 years ago

A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a

OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS A B C

Benzene 1 4.72583 1660.652 -1.461

Toluene 2 4.07827 1343.943 -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5% = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis for all components

So for Benzene ; p × y1=x1 × p1s ------let this be equation 1

for Toluene ; p × y2=x2 × p2s ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0

2 years ago