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Vilka [71]
4 years ago
6

4 The time to failure of an electrical component has a Weibull distribution with parameters λ = 0.056 and a = 2.5. A random coll

ection of 500 components is obtained. Estimate the probability that at least 125 of the 500 components will have failure times larger than 20.

Physics
1 answer:
vivado [14]4 years ago
5 0

Answer:

the probability that at least 125 of the 500 components will have failure times larger than 20 is 0.7939

Explanation:

see the attached file

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padilas [110]
The answer would be C. Gamma Rays and High Frequency EM waves travel at the speed of light and are transverse waves.
7 0
3 years ago
Read 2 more answers
A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

m_c = The center of mass of the stick = \frac{L}{2}-0.22=0.5-0.22=0.28\ m

\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

4 0
3 years ago
Objects that rest have no forces upon them
Pani-rosa [81]

Answer:

false

Explanation:

every object will always have the force of gravity acting upon it.

6 0
3 years ago
A vacuum cleaner has a rating of 460 W on 230 V mains. The value of the fuse connected in the plug will be
amid [387]

Answer:

2 amps

Explanation:

Given data

Power = 460W

voltage= 230V

Required

The amperage/ current of the fuse

Recall  P= IV

I= P/V

I= 460/230

I=2 amps

Hence the current of the fuse is 2 amps

7 0
3 years ago
Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it
notka56 [123]

Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

Extension, e = 4cm to meters = 4/100 = 0.04 meters

Workdone extension = 40cm to meters = 40/100 = 0.4 meters

To find the work done;

First of all, we would find the spring constant using the formula;

Force = spring constant * extension

10 = spring constant * 0.04

Spring constant = 10/0.04

Spring constant = 250 N/m

Next, we find the work done;

Workdone = ½ke²

Where;

k is the spring constant.

e is the extension.

Substituting into the formula, we have;

Workdone = ½ * 250 * 0.4²

Workdone = 125 * 0.16

Workdone = 20 Joules

5 0
3 years ago
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