Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
Answer:
It gives decision-makers the processing capacity they need to turn raw data into useful knowledge.
Explanation:
Data analysis and the associated cycles (data integration, aggregation, hoarding, and revealing) are totally or partially directed in the cloud with cloud analytics.
Answer:
A pointer is spun on a fair wheel of chance having its periphery labeled Trom 0 to 100. (a) Whhat is the sample space for this experiment? (b)What is the probability that the pointer will stop between 20 and 35? (c) What is the probability that the wheel will stop on 58?
Explanation:
thats all you said
Answer: True
Explanation:
Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.
While Engineering strain is the amount that a material deforms per unit length in a tensile test. It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.
