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AlladinOne [14]
2 years ago
14

Explain imprinting in relation to Lorenz's geese. What would most likely have happened if the first thing the goslings saw was a

dog?
Physics
1 answer:
iragen [17]2 years ago
7 0

If the first thing that the goslings saw was a dog, they would have followed the dog as a mother.

Imprinting refers to the process of training an animal to bond with anything it sees after birth even if it is not its real mother. Lorenz first achieved imprinting in 1935 using geese which followed him as their mother shortly after they were born.

If the geese were exposed to a dog, they could also have seen the dog as their mother and followed it accordingly shortly after birth.

Learn more: brainly.com/question/11401513

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Q 2 Two mirrors meet at right angles. A ray of light is incident on one at an angle of 30°
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a ray of light is incident towards a plane mirror at an angles of 30degrees with the mirror surface. what will be the angles of reflection is 60degree.

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3 years ago
Read 2 more answers
2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to
ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b)  the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

     F = m a

    a = F / m

Let's use kinematics to find the velocity after x = 1 m

       v² = v₀² + 2 a x

The initial speed is zero

       v = √ (2 F/m  x)

For the distance of x = 1 m

        v₁ = √ (2 F / m)

Car 2

      F = 2m a

      a = F / 2m

      v² = 2 a x

      v = √ (F/m  x)

 For x = 1 m

       v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

      K₁ = ½ m v₁²

      K₁ = ½ m 2F / m

      K₁ = F

Car 2

      K₂ = ½ 2m v₂²

      K₂ = ½ 2m F / m

      K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

   P₁ = m v₁

   P₁ = m √ (2F / m)

Car 2

    P₂ = 2m v²

    P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

          v = v₀ + a t

          v = F / m t

          v₁ = F / m 10

Car 2

           v₂ = F / 2m 10

           v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

           K₁ = ½ m v₁²

           K₁ = ½ m (F / m 10)²

           K₁ = 50 F² / m

Car2

         K₂ = ½ 2m v₂²

         K₂ = m (F / m 5)²

         K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

         P₁ = m v₁

         P₁ = m F / m 10

         P₁ = 10 F

 

Car 2

        P₂ = 2m v₂

        P₂ = 2m F / m 5

        P₂ = 10 F

In this case the moment of the two cars is the same

7 0
2 years ago
Q.3. The equivalent resistance across AB is:<br> (a)1<br> (c)2<br> (b)3<br> (d)4
sp2606 [1]

Answer:

1 ohm

Explanation:

First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:

1 / Eq = (1 / r₁) + (1 / r₂)

The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:

Eq = r₁ + r₂

Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:

1/E₁ = 1/2 + 1/2

1/E₁ = 1

E₁ = 1Ω

1/E₂ = 1/2 + 1/2

1/E₂ = 1

E₂ = 1Ω

This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:

E₃ = E₁ + E₂ = 1 + 1 = 2Ω

The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:

1/Eq = 1/E₃ + 1/2

1/Eq = 1/2 + 1/2

1/Eq = 1

Eq = 1Ω

7 0
2 years ago
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