Can someone please help, ty!!<br> Will mark brainliest.

Rama09 [41]

Answer:

tan is 15 for that triangle

5 0

2 years ago

For a bronze alloy, the stress at which plastic deformation begins is 294 MPa and the modulus of elasticity is 121 GPa. (a) What

FromTheMoon [43]

Answer:

a) load in Newton is 96,138 b) 129.314mm

Explanation:

Stress = force/ area (cross sectional area of the bronze)

Force(load) = 294*10^6*327*10^-6 = 96138N

b) modulus e = stress/ strain

Strain = stress/ e = (294*10^6)/ (121*10^ 9) = 2.34* 10^ -3

Strain = change in length/ original length = DL/ 129

Change in length DL = 129 * 2.34*10^ -3 = 0.31347

Maximum length = change in length + original length = 129.314mm

7 0

3 years ago

Consider a rifle, which has a mass of 2.44 kg and a bullet which has a mass of 150 grams and is loaded in the firing chamber.Whe

svetlana [45]

Answer:

a. 0 kgm/s

b. 0 kgm/s

c. 66 kgm/s

d. -66 kgm/s

e. 0 kgm/s

f. -27.05 m/s

g. 173.68 N

h. 12.58 m/s

i. 0.772 m

j. 14487 J

Explanation:

150 g = 0.15 kg

a. Before the bullet is fired, both rifle and the bullet has no motion. Therefore, their velocity are 0, and so are their momentum.

b. The combination is also 0 here since the bullet and the rifle have no velocity before firing.

c. After the bullet is fired, the momentum is:

0.15*440 = 66 kgm/s

d. As there's no external force, momentum should be conserved. That means the total momentum of both the bullet and the rifle is 0 after firing. That means the momentum of the rifle is equal and opposite of the bullet, which is -66kgm/s

e. 0 according to law of momentum conservation.

f. Velocity of the rifle is its momentum divided by mass

v = -66 / 2.44 = -27.05 m/s

g. The average force would be the momentum divided by the time

f = -66 / 0.38 = 173.68 N

h. According the momentum conservation the bullet-block system would have the same momentum as before, which is 66kgm/s. As the total mass of the bullet-block is 5.1 + 0.15 = 5.25. The velocity of the combination right after the impact is

66 / 5.25 = 12.58 m/s

i. The normal force and also friction force due to sliding is

$F_f = N\mu = Mg\mu = 5.25*9.81*0.83 = 42.75N$