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Pachacha [2.7K]
3 years ago
13

An air-standard otto cycle has compression ratio 10 and maximum cycle temperature 1000 k. evaluate the thermal efficiency and sp

ecific work of the cycle. what would be the thermal efficiency and specific work if it was a diesel cycle to the same maximum temperature? what would be the thermal efficiency and specific work if it was a dual pressure cycle to the same maximum temperature, where half the energy addition was from constant volume and half the energy addition as from constant pressure?
Physics
1 answer:
Tasya [4]3 years ago
7 0
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A car's gas tank contains 58.7 kg
Bogdan [553]

Answer:

721 kg/m^3

Explanation:

Trust me bro

8 0
3 years ago
Earth has a mass of 5.97 × 1024 kg and a radius of 6.38 × 106 m, while Saturn has a mass of 5.68 × 1026 kg and a radius of 6.03
hammer [34]

Answer: mass on earth = 625.792 N, mass on saturn = 666.75 N

Explanation: weight of an object = mg

Where g = acceleration due to gravity =GM/r²

Where G = gravitational constant, M = mass of planet and r = radius of planet.

Let us start with earth.

G =gravitational constant of the earth =6.67×10^-11 m³ /kgs²

Me = mass of earth = 5.97×10^24 kg

re = radius of earth = 6.38×10^6 m

Acceleration due to gravity on earth = (6.67×10^-11×5.97×10^24) /(6.38×10^6)²

Acceleration due to gravity on earth = 3.98×10^14 /4.07×10^13

Acceleration due gravity on earth = 9.778 m/s²

The mass of the person is 64 kg, hence his weight on earth is given as

W = mass × Acceleration due gravity on earth = 64 ×9.778 = 625.792 N

For saturn.

G =gravitational constant of the earth =6.67×10^-11 m³ /kgs²

Ms = mass of saturn = 5.68×10^26kg

rs= radius of saturn= 6.03×10^7 m

Acceleration due to gravity on saturn= (6.67×10^-11×5.68×10^26 /(6.03×10^7)²

Acceleration due to gravity on saturn= 3.788×10^16/3.636×10^15

Acceleration due gravity on saturn= 10.417 m/s²

The mass of the person is 64 kg, hence his weight on saturn is given as

W = mass × Acceleration due gravity on saturn = 64 ×10.417 = 666.75 N

3 0
3 years ago
A rifle is aimed horizontally at a target 47 m away. The bullet hits the target 2.3 cm below the aim point.
inna [77]

Answer:

Is your question asking for the muzzle velocity of the bullet?

Explanation:

I will assume it does

The bullet travels horizontally to the target in the same amount of time it falls 2.3 cm from vertical rest

s = ½at²

t = √(2s/g) = √(2(0.023) / 9.8) = 0.0685118...s

v = d/t = 47/0.0685118 = 686.01242...

v = 690 m/s

3 0
2 years ago
What hormone is not related to stress
viktelen [127]
The answer would be Estrogen, because that is a female sex hormone
8 0
4 years ago
A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s
Evgen [1.6K]

Answer:

(a) I (Moment of inertia)=0.0987 kgm^{2}

(b) W(Angular Speed)=2.66 \frac{rad}{s}

Explanation:

Given data

m (Monkey mass)= 1.80 kg

d=2.50 m

T (Time Period)=0.940 s

Angle= 0.400 rad

(a) I (Moment of Inertia)=?

(b) W (Angular Speed)=?

For part (a) I (Moment of Inertia)=?

Time Period Formula is given as

T=2(3.14)\sqrt{\frac{I}{mgd} }

After Simplifying we get

I=\frac{mgdT^{2} }{4*(3.14)^{2}}

I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }

I=0.0987 kgm^{2}

For Part (b) Angular Speed

From Kinetic Energy we get

KE=\frac{1}{2}IW^{2}

Pontential Energy

PE=mgd(1-Cosa)

KE=PE

\frac{1}{2}IW^{2}=mgd(1-Cosa)

W^{2}=\frac{2mgd(1-Cosa)}{I}

W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }

W=2.66\frac{rad}{s}

5 0
3 years ago
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