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11111nata11111 [884]
3 years ago
15

In proof testing of circuit boards, the probability that any particular diode will fail is 0.01. Suppose a circuit board contain

s 200 diodes.
(a) How many diodes would you expect to fail?
2 diodes
Physics
1 answer:
Sergio039 [100]3 years ago
7 0

Answer: 2 diodes

Explanation:

Given : In proof testing of circuit boards, the probability that any particular diode will fail is 0.01.

The number of diodes contained by circuit board = 200

Then , the expected number of diodes to fail is given by :-

0.01\times200=2

Therefore, there are 2 diodes that we will expect to fail.

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Which of the following are not vector directions?
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here north are not vector option b hope ur help

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As the boat in which he is riding approaches a dock at 3.0 m/s, Jasper stands up in the boat and jumps toward the dock. Jasper a
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Im pretty sure it’s a because it makes more sense you know?.
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A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

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Using conservation of energy

K.E+P.E=P.E+K.E

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\dfrac{v_{1}^2}{2}=18.62

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Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

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