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joja [24]
2 years ago
12

You look in the sky and see two jetliners that you know are of equal size, yet one appears to be much larger. Because of your kn

owledge of __________, you will assume that the smaller jetliner is farther away
Physics
1 answer:
Ksju [112]2 years ago
8 0

Answer:

Because of the knowledge of <u>relative size</u>, it will be assumed that the smaller jetliner is farther away.

Explanation:

According to the theory of relative size, the distance that an object has to the viewing individual affects the perception of the individual regarding the size of the object.

As stated in this case, one of the jetliners is farther away from the other. Therefore, even if the jets are of equal size, the one that is at a greater distance is perceived to be smaller as it is at a greater viewing range. The one that is nearer to the individual seems bigger in comparison to the one farther away due to a closer viewing range.

Therefore, the jet that is nearer appears larger.

To know more about relative size, refer to:

brainly.com/question/19998265

#SPJ4

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What is the
Len [333]

Answer:

d=5\ g/cm^3

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3

So, the density of the object is equal to 5\ g/cm^3.

6 0
3 years ago
A light spring stretches 0.13 m when a 0.35 kg mass is hung from it. The mass is pulled down from this equilibrium position an a
Alenkasestr [34]

Answer:

v = 1.30 m/s

Explanation:

given,

mass hung = 0.35 Kg

spring stretched when load is hanged  (x)= 0.13 m

now,

weight of the mass attached = Kx

             m g = k x

             0.35 x 9.8 = k x 0.13

                k = 26.38 N/m

now, using conservation of energy

 \dfrac{1}{2}mv^2 = \dfrac{1}{2}kx'^2

 v = \sqrt{\dfrac{kx'^2}{m}}

 v = \sqrt{\dfrac{26.38 \times 0.15^2}{0.35}}

 v = \sqrt{1.6958}

    v = 1.30 m/s

6 0
3 years ago
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
3 years ago
8)
Jobisdone [24]
Hey!!

here is your answer >>>

We know that, acids and strong bases are good conductors of electricity!. And gold conducts electricity , while the best is silver and niext comes gold!. And H20 which is water contains less ions , so it is not a good conductors , it is a conductors but not a good conductor!. Salt in it's solid for cannot produce ions, hence, it cannot conduct electricity , but when this salt is mixed with water or when it is in it's molten state it can conduct electricity!. So now,the answer for the question is ,

B) III and IV only!.

Hope my answer helps!
7 0
3 years ago
Read 2 more answers
A 12.0 khz, 16.0 v source connected to an inductor produces a 4.00 a current. what is the inductance?
Varvara68 [4.7K]
Inductive reactance (Z) = ω L  =  2Πf L = (2Π) (12,000) (L)

I = V / Z

4 A = 16v / (24,000Π L)

Multiply each side by (24,000 Π L):

96,000 Π L = 16v

Divide each side by  (96,000 Π) :

L = 16 / 96,000Π  =  5.305 x 10⁻⁵ Henry

L = 53.05 microHenry
4 0
3 years ago
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