Answer:
longitudinal engineering strain = 624.16
true strain is 6.44
Explanation:
given data
diameter d1 = 0.5 mm
diameter d2 = 25 mm
to find out
longitudinal engineering and true strains
solution
we know both the volume is same
so
volume 1 = volume 2
A×L(1) = A×L(2)
( π/4 × d1² )×L(1) = ( π/4 × d2² )×L(2)
( π/4 × 0.5² )×L(1) = ( π/4 × 25² )×L(2)
0.1963 ×L(1) = 122.71 ×L(2)
L(1) / L(2) = 122.71 / 0.1963 = 625.16
and we know longitudinal engineering strain is
longitudinal engineering strain = L(1) / L(2) - 1
longitudinal engineering strain = 625.16 - 1
longitudinal engineering strain = 624.16
and
true strain is
true strain = ln ( L(1) / L(2))
true strain = ln ( 625.16)
true strain is 6.44
As we know that range of the projectile motion is given by

here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be

so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree
It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.
Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.
Answer:
Final speed of car = 12 m/s
Explanation:
We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.
a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s
v = ?
u = 0 m/s
a = 4.0 m/s²
t = 5 s
v = u + at = 0 + 4 x 5 = 20 m/s
b) Then maintains that velocity for 10 s
v = ?
u = 20 m/s
a = 0 m/s²
t = 10 s
v = u + at = 20 + 0 x 10 = 20 m/s
c) Then decelerates at the rate of 2.0 m/s² for 4.0 s
v = ?
u = 20 m/s
a = -2.0 m/s²
t = 4 s
v = u + at = 20 + -2 x 4 = 12 m/s
Final speed of car = 12 m/s
The random variable in this experiment is a Continuous random variable.
Option D
<u>Explanation</u>:
The continuous random variable is random variable where the data can take infinite variables. For example random variable is taken for measuring "speed of automobiles" on the highways. The radar instrument depicts time taken by automobile in particular what speed. They are the generalization of discrete random variables not the real numbers as a random data is created. It gives infinite sets of all possible outcomes. It is obvious that outcomes of the instrument depend on some "physical variables" those are not predictable as depends on the situation.