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3 years ago

Select the correct answer from each drop-down menu.

Physics

1 answer:

earnstyle [38]3 years ago

7 0

Answer:

a transverse (sort of a plot of a sine or cosine graph, basically)

b longitudinal

c Electromagnetic (an electric wave and a magnetic wave travelling together at right angles to each other)

Explanation:

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Carbon dioxide enters an adiabatic nozzle steadily at 1 MPa, 518 oC, and mass flow rate of 5,322 kg/h and exits the system at 96

Orlov [11]

Answer:

The velocity at the nozzle at inlet $V_{1}$ = 3584 $\frac{m}{sec}$

Explanation:

Pressure at inlet $P_{1}$ = 1 × $10^{6}$ Pa

Temperature at inlet $T_{1}$ = 518 ° c = 791 K

Mass flow rate = $\frac{5322}{60}$ $\frac{kg}{sec}$ = 88.7

Gas constant for carbon die oxide is R = 189 $\frac{J}{kg k}$

Mass flow rate inside the nozzle is given by the formula = $\frac{P_{1} }{R T_{1} }$ × $A_{1}$ × $V_{1}$

⇒ $P_{1}$ = = 1 × $10^{6}$ Pa

⇒ R $T_{1}$ = 791 × 189 = 149499 $\frac{J}{kg}$

⇒ $A_{1}$ = 0.0037 $m^{2}$

Put all the above values in above formula we get,

⇒ 88.7 = $\frac{10^{6} }{149499}$ × 0.0037 × $V_{1}$

⇒ $V_{1}$ = 3584 $\frac{m}{sec}$

This is the velocity at the nozzle at inlet.

3 0

3 years ago

Sound travels at a rate of 340 m/s in all directions through air. Matt rings a very loud bell at one location, and Steve hears i

Vika [28.1K]

Answer:

110 m/s

Explanation:

because if you subtract 450 from 340 you get 110

6 0

2 years ago

The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad

Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

$A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})$

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

$A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )$

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

T = 2*p / w

T = 2*p / (2*p / 92.8)

Hence, T = 92.8 min

7 0

3 years ago

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

ad-work [718]

Answer:

Maximum speed of the car is 17.37 m/s.

Explanation:

Given that,

Radius of the circular track, r = 79 m

The coefficient of friction, $\mu=0.39$

To find,

The maximum speed of car.

Solution,

Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :

$v=\sqrt{\mu rg}$

$v=\sqrt{0.39\times 79\times 9.8}$

v = 17.37 m/s

So, the maximum speed of the car is 17.37 m/s.

6 0

3 years ago

Calculate the potentia energy of a car with a mass of 3800kg that is on a hill 110 meters above sea level? USE 10 instead of 9.8

Arada [10]

Potential energy (PE ) = m g h

Where:

m = mass = 3800 kg

g = acceleration due gravity = 10 m/s^2

h = heigth = 110 meters

Replacing:

PE = 3800 * 10 * 110 = 4,180,000 J

7 0

1 year ago