Paul is driving on a sunny day what glasses should he wear

A pure jet engine propels an aircraft at 240 m/s through air at 45 kPa and −13°C. The inlet diameter of this engine is 1.6 m, th

erica [24]

QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.

ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.

6 0

3 years ago

Which of the following statements about the conservation of momentum is not correct? Momentum is conserved for a system of objec

Olin [163]

Answer:

wrong statement : Momentum is not conserved for a system of objects in a head-on collision.

Explanation:

In a head on collision of two objects , two equal and opposite forces are created at the point of collision . These two forces create two impulses in opposite direction which results in equal and opposite changes in momentum in each of them . Hence net change in momentum is zero. In this way momentum is conserved in head on collision of two objects.

3 0

3 years ago

A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

3 years ago

A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

A(n) 96.1 g ball is dropped from a height of 59.1 cm above a spring of negligible mass.The ball compresses the spring to a maxim

Serhud [2]

Answer:

Explanation:

Mass of ball Is m=96.1g=0.0961kg

Height above spring is 59.1cm

L=0.591m

Extension of the spring is 4.75403cm

e=0.0475403m

Then the distance the ball traveled is H=L+e

H=0.591+0.0475403

H=0.6385403m

Then, the potential energy of the ball is given as

P.E=mgh

P.E=0.0961×9.81×0.6385403

P.E=0.602J

From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another

Then, the P.E is transferred to the work done by the spring

Then, Work done by spring is given as

W=½ke²

W=P.E=½×k×0.0475403²

0.602=½×k×0.0475403²

k=0.602×2/0.0475403²

k=532.72N/m

The spring constant is 532.72 N/m

4 0

3 years ago