The collapse of the magnetic field inside the ignition coil happens as a

What is the magnetic force on a moving electric charge called

crimeas [40]

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule. The force is given by the charge times the vector product of velocity and magnetic field.

4 0

3 years ago

Read 2 more answers

What are the indicators of ineffective systems engineering?

liberstina [14]

Answer:

Indicators for ineffective system engineering are as follows

1.Requirement trends

2.System definition change backlog trends

3.interface trends

4.Requirement validation trends

5.Requirement verification trends

6.Work product approval trends

7.Review action closure trends

8.Risk exposure trends

9.Risk handling trends

10.Technology maturity trends

11.Technical measurement trends

12.System engineering skills trends

13.Process compliance trends

7 0

3 years ago

.

Gennadij [26K]

Answer:

II

Explanation:

3 0

3 years ago

Calculate the magnitude of the velocity and the θ angular direction of the block and the bullet together when the 50 g bullet mo

almond37 [142]

Answer:

Magnitude of the velocity = 16.82 m/s

Angular direction, θ = 52.41°

Explanation:

As given ,

mass of bullet, m₁= 50g = 0.05 kg

speed of bullet , u₁ = 600 m/s

mass of the block , m₂ = 4 kg

speed of the block before collision , u₂ = 12 m/s

direction , θ = 30°

Now,

Assume that the combined velocity of bullet and block after collision = v

and the direction = θ

Now, from the conservation of momentum in x - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) vₓ

where v = final velocity after collision

u₁ = initial velocity of bullet before collision = 0

m₁ = mass of the bullet before collision = 0.05 kg

u₂ = velocity of block before collision = 12 cos(30° )

m₂ = mass of block before collision

m₁ + m₂ = combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (0) + 4(12 cos(30° ) ) = ( 0.05 + 4 ) vₓ

⇒ 0 + 4(6√3) = 4.05 vₓ

⇒24√3 = 4.05 vₓ

⇒vₓ = 10.26 m/s

Now, from the conservation of momentum in y - direction :

m₁ u₁ + m₂ u₂ = ( m₁ + m₂ ) $v_{y}$

where v = final velocity after collision

u₁= initial velocity of bullet before collision = 600

m₁ = mass of the bullet before collision = 0.05 kg

u₂ = velocity of block before collision = 12 sin(30° )

m₂= mass of block before collision

m₁+ m₂= combined mass of bullet and block after collision = 0.05 + 4

∴ we get

0.05 (600) + 4(12 sin(30° ) ) = ( 0.05 + 4 ) $v_{y}$

⇒ 30 + 4(6) = 4.05 $v_{y}$

⇒30 +24 = 4.05 $v_{y}$

⇒54 = 4.05 $v_{y}$

⇒ $v_{y}$ = 13.33 m/s

Now, the magnitude of the velocity = √vₓ² + $v_{y}$ ² = √(10.26)² + (13.33)²

= √105.26 + 177.68

= √282.95 = 16.82

The angular direction, θ = $tan^{-1}$ ( $\frac{v_{y} }{v_{x} }$ ) = $tan^{-1}$ ( $\frac{13.33}{10.26}$ ) = $tan^{-1}$ (1.299) = 52.41°

8 0

3 years ago

A liquid propellant engine has the following characteristics: chamber pressure of 7 MPa, constant ratio of specific heats of 1.3

Cloud [144]

Answer:

1.55
260 N.s
3370 m
1.6
43.75 kg/s

Explanation:

1) Thrust coefficient at sea level.

Cfsl = TSL / Pca

TSL = Mp * Vc + ( Pc - Pa )Ac

Mp = mass flux = 43.75 kg/s

∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )

= 1.6 - 0.04923 = 1.55

2) Specific impulse at sea

Isp = Vc / g = 2549.75 / 9.81

= 260 N.s

3) Altitude at optimal expansion

H = 3370 m

4) thrust coefficient at optimal expansion

CF = 1.6

attached below is the detailed solution

5) Mass flux through the throat

Mass flux = P1 * At / Cc

= ( 7*10^6 * 0.01 ) / 1600

= 43.75 kg/s

8 0

3 years ago