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mylen [45]
3 years ago
13

A truck contains both hydraulic brakes (reliability 0.96) and mechanical brakes (reliability 0.99). What is the probability of s

topping the truck (i) while moving at a high speed, assuming both systems must work to avoid hitting a vehicle stalled in the roadway, and (ii) while moving at a low speed, assuming either system will stop the truck in time if they function properly
Physics
1 answer:
Oksana_A [137]3 years ago
8 0

Answer:

i) 0.9504

ii) 0.0452

Explanation:

Given data: reliability of hydraulic brakes= 0.96

reliability of mechanical brakes = 0.99

So the probability of stopping the truck = 0.96×0.99= 0.9504

At low speed

case: A works and B does not

= 0.96×(1-0.99) = 0.0096

case2 : B works and A does not

= 0.99×(1-0.96) = 0.0396

Therefore, probality of stopping = 0.0096+0.0396 = 0.0492

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Answer:

3 m/s^2

Explanation:

acceleration= Change in velocity/time

= 30-0 / 10

= 30/10

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Please help! Answer and explain
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Answer:

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Which statement correctly describes the relationship between frequency and wavelength?
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The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency. 

If we divide the equation by f we get:

λ=v/f

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So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.  
3 0
3 years ago
Read 2 more answers
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
What is the length of a spring that has 450J of potential energy and a spring constant of 650N/m?
11111nata11111 [884]

Answer:

Δx = 1.2 m

Explanation:

The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²

Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m

The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.

4 0
3 years ago
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