Answer:
a) —0.5 j m/s
b) 4.5 i + 2.25 j m
Explanation:
<u>Givens:</u>
v_0 =3.00 i m/s
a= (-3 i — 1.400 j ) m/s^2
The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :
<em>v_x = v_0 + at = 0 </em>
(3.00 i m/s) + (-3 i m/s^2)t=0
t = (3 m/s)/-3 i m/s^2
t = -1 s
Therefore the particle reaches the maximum x-coordinate at time t = 1 s.
Part a The velocity-of course- is all in the y-direction,therefore:
v_y =v_0+ at
We have that v_0 = 0 in the y-direction.
v_y = (-0.5 j m/s^2)(1 s)
= —0.5 j m/s
Part b: While the position of the particle at t = 1 s is given by:
r=r_0+v_0*t+1/2*a*t^2
Where r_0 = 0 since the particle started from the origin.
Its position at t = 1 s is then given by :
r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2
=4.5 i + 2.25 j m
Answer:
The correct answer is (a), heating and soaking at a temperature above the austenitizing level followed by rapid cooling.
Explanation:
Tempering is a heat treatment that is performed to change microstructure of steel <em>and increase it's hardness.</em>
It is done by heating above the austenitizing level, because for this heat treatment it's necesary to heat the steel to get <em>austenite structure</em> that is the <em>alotropic form of gamma iron.</em>
The procedure of tempering, continues with rapid cooling, because this way you get the microstructure with the best hardness properties, that is <em>martensite.</em>
I can't see it maybe take another picture?
Answer:
distance is 139.2 cm
image is inverted and virtual
and 79.36 cm in front of 2nd lens
Explanation:
Given data
convex focal length = 24 cm
distance = 50 cm
concave focal length = -42 cm
light bulb height = 2 cm
distance = 29 cm
to find out
Determine the distance and orientation and image real or virtual
solution
we apply here lens formula for convex
1/f = 1/p + 1/q
here f = 24 and p = 29 so find q
1/24 = 1/29 + 1/q
q = 139.2 cm
so distance is 139.2 cm
and
lens formula for concave
f = - 42 and p = 50 - 139.2 = -89.2 cm so find q
1/f = 1/p + 1/q
- 1/42 = - 1/89.2 + 1/q
q = −79.36 cm
so image is inverted and virtual
and 79.36 cm in front of 2nd lens
135 is the best way to get to the acceleration
