Answer:
3rd and 4rth
Explanation:
a geologist studies the earth and both of these have something to do with the earth
The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².
<h3>What is normal stress?</h3>
If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.
σ = P/A
Where, σ = Normal stress
P = Pressure
A = Area
1 Kg = 9.81 N
800 kg = 7848 N
Since the rod is half bronze and half steel
800 kg = 7848/2
= 3924 N
Pₙ = Fₙ = 3924 N [n = Bronze]
Pₓ = 3924 N [x = steel]
Given,
σₙ = 90MPa
σₓ = 120MPa
Aₙ = ?
Aₓ = ?
Aₙ = Pₙ/σₙ
Aₙ = 3924/90
Aₙ = 43.6 mm²
Aₓ = Pₓ/σₓ
Aₓ = 3924/120
Aₓ = 32.7 mm²
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Answer:
64.11% for 200 days.
t=67.74 days for R=95%.
t=97.2 days for R=90%.
Explanation:
Given that
β=2
Characteristics life(scale parameter α)=300 days
We know that Reliability function for Weibull distribution is given as follows

Given that t= 200 days

R(200)=0.6411
So the reliability at 200 days 64.11%.
When R=95 %

by solving above equation t=67.74 days
When R=90 %

by solving above equation t=97.2 days
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Answer:


Explanation:
Given that:
x(t) = 10 sin(10t) . sin (15t)
the objective is to find the power and the rms value of the following signal square.
Recall that:
sin (A + B) + sin(A - B) = 2 sin A.cos B
x(t) = 10 sin(15t) . cos (10t)
x(t) = 5(2 sin (15t). cos (10t))
x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)
x(t) = 5sin(25 t) + 5 sin (5t)
From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:

For the number of sinosoidial signals;
Power can be expressed as:

As such,
For x(t), Power 



For the number of sinosoidial signals;

For x(t), the RMS value is as follows:




