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otez555 [7]
3 years ago
15

Plz help electrical technology

Engineering
2 answers:
oksano4ka [1.4K]3 years ago
8 0

Answer:

OPTION A,Larger

HOPE IT HELPS

ozzi3 years ago
3 0
The answer should be option A. Larger
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Cuál de las siguientes es la mejor manera de practicar sus habilidades de tecnología de secundaria?
Mkey [24]
Huh? Do you know English?
8 0
3 years ago
Information such as tolerances and scale can be found in the _______________ of an engineering drawing.
nasty-shy [4]

Answer:

Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing

Explanation:

The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item

8 0
2 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
2 years ago
Thế nào là góc quay xe ?
slava [35]

Answer:

what's what is what is it return if you have any clue please let me know Translate in English and send me because most of them are are from English I think so

8 0
2 years ago
Which ranks careers in the Architecture and Construction career cluster based on the typical number of years for each degree, go
timama [110]

Answer:

hmm tdyufrjhfjjbdjjkljgjjddtu

8 0
3 years ago
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