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Temka [501]
3 years ago
11

A __________ is a push or pull.​

Physics
1 answer:
maria [59]3 years ago
5 0

Answer:

force

Explanation:

Force is a push or a pull of an object that causes the object to speed up, slow down, or stay in one place. In other words, a force is what causes an object to move. Friction and gravity are two types of forces that influence how an object moves.

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(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet
trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

\dot m = \frac{70000}{510.88 - 311.73}

\dot m = 351.49 kg/s

b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

\dot m_{actual} = 1046.15 kg/s

6 0
3 years ago
Question 24
Verizon [17]
Q 24: Newton's first law of motion states that anything that stands still will stay still unless unbalanced force is acted upon that object. Newton's second law of motion states that the velocity of an object changes when subjected to external Force. Newton's third law of motion says that for every action there is an equal and opposite reaction. when bumper cars crash into each other one car goes one way and the other goes the other way.

Q 25: if the bumper cars doubled their Mass the motion would be halfed but if the net force is doubled the motion will double.
8 0
3 years ago
What is a discussion?
kap26 [50]

Answer:

um, when you talk with other people about stuff. (I'm not trying to sound like a smarta s s I'm just giving a definition...)

Explanation:

7 0
3 years ago
Read 2 more answers
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.​
docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

7 0
3 years ago
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s
nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where:

F=1.1 \cdot 10^5 N is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

I=\Delta p = m \Delta v

where

m = 68 kg is the mass of the man

\Delta v = 0-55 m/s = -55 m/s is the change in velocity of the man

Substituting,

I=(68 kg)(-55 m/s)=-3740 kg m/s

7 0
3 years ago
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