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Sedbober [7]
3 years ago
7

3

Physics
1 answer:
Nitella [24]3 years ago
8 0

Answer:

The displacement is 6.4m

Explanation:

Step one:

given

we are told that Lucy runs 4 meters to the east,

then 5 meters south.

let the distance east be the displacement in the x-direction, and south be the y-direction

Step two:

The resultant of the x and y displacement is the magnitude of the total displacement z

applying Pythagoras theorem we have

z=√x^2+y^2

z=√4^2+5^2

z=√16+25

z=√41

z=6.4m

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A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo
guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s (1km\h=\frac{5}{18} m/s)

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)

a = 1.83 m/s^2

Force=mass\times acceleration

Force=710 kg\times 1.83 m/s^2

Force=1,299.3 N

The net force acting on the car 1,299.3 N.

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A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

                           = T/2 = 0.952/2 = 0.476 kips/in²

8 0
3 years ago
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