Answer: all the above options are correct.
Explanation:
In sidewall markings,the load index is given as a letter,traction and temperature ratings are based on the speed rating of the tire,the tire's recommended inflation pressure and load are indicated and the DOT code indicates when and where the tire was made.
Answer:
a)
b)
Explanation:
Given:
mass of bullet, 
compression of the spring, 
force required for the given compression, 
(a)
We know

where:
a= acceleration


we have:
initial velocity,
Using the eq. of motion:

where:
v= final velocity after the separation of spring with the bullet.


(b)
Now, in vertical direction we take the above velocity as the initial velocity "u"
so,

∵At maximum height the final velocity will be zero

Using the equation of motion:

where:
h= height
g= acceleration due to gravity


is the height from the release position of the spring.
So, the height from the latched position be:



Contact metamorphism occurs adjacent to igneous intrusions and results from high temperatures associated with the igneous intrusion. Since only a small area surrounding the intrusion is heated by the magma, metamorphism is restricted to the zone surrounding the intrusion, called a metamorphic or contact aureole
Answer:
15 m/s
Explanation:
Speed(m/s) = distance(m)/time(s)
distance = 216 km = 216,000 m
time = 4 hours = 14,400 s
speed = 216000/14400 = 15 m/s