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alex41 [277]
3 years ago
9

Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 7.50 cm apart. You see an object jump from sid

e to side by 0.95 degree as you blink back and forth between your eyes. How far away is the object?
Physics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

Distance of the object from eye is approx 4.52 m

Explanation:

As we know that the object subtend a small angle on both the eyes which is given as

\theta = 0.995 degree

now we know that the distance between two eyes is given as

d = 7.50 cm

so we have

angle = \frac{arc}{Radius}

so here the radius is same as the distance from eye while arc is the distance between two eyes

so we have

0.95 \frac{\pi}{180} = \frac{7.50}{R}

R = 452 cm

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3 years ago
Question: A car (assumed to be a Ford Taurus) is traveling around a turn that is banked at 7 degrees. The turn has a radius of 2
AysviL [449]

Answer:

Question: A car (assumed to be a Ford Taurus) is traveling around a turn that is banked at 7 degrees. The turn has a radius of 29 m. The car has a mass of 1300 kg. The coefficient of static friction between the tires and the road is 0.68.

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2 years ago
Air expands adiabatically in a piston–cylinder assembly from an initial state where p1 = 100 lbf/in.2, v1 = 3.704 ft3/lb, and T1
lutik1710 [3]

Answer:

Final temperature is equal to 1291.63°R  

Explanation:

given,

p₁ = 100 lb f/in²,               v₁ = 3.704 ft³/lb,           and T₁ = 1000 °R

p₂ = 30 lb f/in²                 n = 1.4

Δ u = 0.171(T₂ - T₁)

we know for poly tropic process

p vⁿ = constant

p₁ v₁ⁿ = p₂ v₂ⁿ

100 × 3.704¹°⁴ = 30 × v₂¹°⁴

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work done for poly tropic process

W = \dfrac{p_1v_1-p_2v_2}{n-1}

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    = 269.525 lbf/in².ft³

W = \dfrac{269.525}{5.40395} Btu/lb

   = 49.87 Btu/lb

in the piston cylinder arrangement air is expanding acrobatically

Δ q = Δu + w

Δ u = - w

0.171(T₂ - T₁) = -49.87

0.171(T₁ - T₂) = -49.87

0.171 T₂ = 0.171 × 1000 + 49.87

T₂ = 1291.63 °R

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5 0
2 years ago
Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st
DedPeter [7]

Answer:

E1  = 10.15 * 10^4 N/C

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Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q)  = +11nC

distance between thin glass rods   = 4 cm .

<u>Calculate the electric field strengths </u>

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applying equation 1 to determine E1

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