Given:
applied tensile stress, = 170 MPa
radius of curvature of crack tip, = mm
crack length = mm
half of internal crack length, a = <u> </u>
a =
Formula Used:
Solution:
Using the given formula:
= 2726 MPa (395372.9 psi)
Answer:
Explanation:
Following relation exists between number of turns and volt
Volt in primary / volt in secondary = no of turns in primary / no of turns in sec
19000 / 120 = no of turns in primary / 159
no of turns in primary = 19000 x 159 / 120
= 25175 Ans .
Answer:
1. 19.28 secs
2. 154.22 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 16 m/s
Final velocity (v) = 0
Force (F) = 1000 N
Mass (m) = 1200 Kg
Time (t) =..?
Distance (s) =...?
Next, we shall determine the acceleration of the car. This can be obtained as follow:
Force (F) = 1000 N
Mass (m) = 1200 Kg
Acceleration (a) =.?
Force (F) = mass (m) x acceleration (a)
F = ma
1000 = 1200 x a
Divide both side by 1200
a = 1000/1200
a = 0.83 m/s²
Since the car is coming to rest, it means it is decelerating. Therefore, the acceleration is – 0.83 m/s²
1. Determination of time taken for the car to halt i.e stop. This can be obtained as follow:
Initial velocity (u) = 16 m/s
Final velocity (v) = 0
acceleration (a) = – 0.83 m/s²
Time (t) =.?
v = u + at
0 = 16 + (–0.83 x t)
0 = 16 – 0.83t
Rearrange
0.83t = 16
Divide both side by 0.83
t = 16/0.83
t = 19.28 secs.
Therefore, the time taken for the car to halt is 19.28 secs.
2. Determination of the distance travelled by the car before coming to rest. This can be obtained as follow:
Initial velocity (u) = 16 m/s
Final velocity (v) = 0
acceleration (a) = – 0.83 m/s²
Distance (s) =..?
v² = u² + 2as
0 = 16² + (2 x –0.83 x s)
0 = 256 – 1.66s
Rearrange
1.66s = 256
Divide both side by 1.66
s = 256/1.66
s = 154.22 m
Therefore, the distance travelled by the car before coming to rest is 154.22 m.
Answer:
A small force is applied over a long time interval can produce a large change in the object's momentum
Explanation:
Answer:
Explanation:
The figure is not provided and we don't know the force applied to the books.
Therefore, here we are going to assume the force applied is 3 N:
F = 3 N
We can find the acceleration of the stack of books by using Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:
where
F is the force
m is the mass
a is the acceleration
In this problem we have
m = 1.5 kg is the mass of the books
Therefore, the acceleration is: