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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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A force is directly proportional to what ?
pishuonlain [190]
It's mass and acceleration
4 0
3 years ago
If a rooster lays and egg on the middle of the roof, which way does the egg role?
Mekhanik [1.2K]

well it depends on the wind and gravity and the side the chickens on and the amount of force but if its a simple question then it doesnt roll at all or it rolls forward

8 0
3 years ago
The angle of reflection is the angle between the normal line and the___?
Aleks [24]

Answer:

Reflected line

Explanation:

The angle between the reflected ray and the normal ray is called angle of reflection.

4 0
3 years ago
A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =
slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

7 0
3 years ago
The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic
Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight =   \frac{2uSin(A)}{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = \frac{2(4)Sin(60)}{10} = 0.7

ii) distance travel = Range =  R = \frac{u^{2}Sin(2A) }{g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = \frac{4^{2}Sin(2*60) }{10} = 1.39

iii) Maximum Height = H = \frac{u^{2}(Sin(A))^{2}  }{2g}

where u = speed = 4, A = 60 and  g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: \frac{4^{2}(Sin60)^{2}  }{2*10} = 0.6

4 0
3 years ago
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