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poizon [28]
3 years ago
15

Why do you think signal detection is a vital concern?

Physics
1 answer:
vazorg [7]3 years ago
7 0

Answer:

Signal Detection Theory, or SDT, is probably the most important and influential theoretical framework for modeling how perceptual decisions are made in forced-choice tasks. Its central concept is that perceptual decisions are based on a stochastic or probabilistic representation of stimuli inside the brain.

Explanation:

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Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
9. A car driver brakes gently. Her car slows down front --
sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

b)

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0
3 years ago
Mention one life application on density ​
kow [346]

One well-known application of density is determining whether or not an object will float on water. If the object's density is less than the density of water, it will float; if its density is less than that of water, it will sink.In fact, submarines dive below the surface of the water by emptying their ballast tanks

5 0
2 years ago
If an object is thrown upward with an initial velocity of 128 ​ft/second, then its height after t seconds is given by the follow
IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

Initial velocity u= 128 ft/sec

Equation of height

h = 128t-32t^2....(I)

(a). We need to calculate the maximum height

Firstly we need to calculate the time

\dfrac{dh}{dt}=0

From equation (I)

\dfrac{dh}{dt}=128-64t

128-64t=0

t=\dfrac{128}{64}

t=2\ sec

Now, for maximum height

Put the value of t in equation (I)

h =128\times2-32\times4

h=128\ ft

(b). The number of seconds it takes the object to hit the ground.

We know that, when the object reaches ground the height becomes zero

128t-32t^2=0

t(128-32t)=0

128=32t

t=4\ sec

Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

3 0
3 years ago
In the following diagram, the voltage is 1.5 volts and the resistance is 0.35 ohms. Use Ohm's Law to determine the current in th
Angelina_Jolie [31]

Answer:

I = 4.28 [amp]

Explanation:

To solve this type of problems we must have knowledge of the law of ohm, which tells us that the voltage is equal to the product of resistance by current.

Initial data:

v = 1.5 [volt]

R = 0.35 [ohms]

v = I * R

therefore:

I = 1.5 / 0.35

I = 4.28 [amp]

5 0
2 years ago
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