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nadezda [96]
3 years ago
9

A person uses their hand to lift a 0.05-kg guniea pig from the ground at a constand speed.What is the force applied by the perso

n on the guinea pig? What is the net force on the guinea pig?
Physics
1 answer:
Ostrovityanka [42]3 years ago
3 0
The animal's weight is

                         (mass) x (gravity)

                 =      (0.05 kg) x (9.8 m/s²)  =  0.49 newton.

That's the minimum force you can use to pick it up.

That's the force applied by the person to pick up the pig
at a nice gentle constant speed.

Then the net force on the pig is zero.  The upward and downward
forces add up to zero, and that's why the animal is moving at a
constant speed, and not accelerating up or down.
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Consider the interference pattern produced by two parallel slits of width a and separation d, in which d = 3a. The slits are ill
laila [671]

Answer:

a)   m =1  θ = sin⁻¹  λ  / d,  m = 2        θ = sin⁻¹ ( λ  / 2d) ,   c)     m = 3

Explanation:

a) In the interference phenomenon the maxima are given by the expression

         d sin θ = m λ

the maximum for m = 1 is at the angle

          θ = sin⁻¹  λ  / d

the second maximum m = 2

          θ = sin⁻¹ ( λ  / 2d)

the third maximum m = 3

        θ = sin⁻¹ ( λ  / 3d)

the fourth maximum m = 4

       θ = sin⁻¹ ( λ  / 4d)

b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.

       I = I₀ cos² (Ф) (sin x / x)²

       Ф = π d sin θ /λ

       x = pi a sin θ /λ

where a is the width of the slits

with the values ​​of part a are introduced in the expression and we can calculate intensity of each maximum

c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present

maximum interference       d sin θ = m λ

first diffraction minimum    a sin θ = λ

we divide the two expressions

                       d / a = m

In our case

                   3a / a = m

                    m = 3

order three is no longer visible

7 0
3 years ago
A ship 1200m off shore fires a gun. how long after the gun is fired will it be heard on the shore?​
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Answer:

We know that the speed of sound is 343 m/s in air

we are also given the distance of the boat from the shore

From the provided data, we can easily find the time taken by the sound to reach the shore using the second equation of motion

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since the acceleration of sound is 0:

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Replacing the variables in the equation with the values we know

1200 = 343 * t

t = 1200 / 343

t = 3.5 seconds (approx)

Therefore, the sound of the gun will be heard at the shore, 3.5 seconds after being fired

6 0
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