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shutvik [7]
3 years ago
13

The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli

ght (wavelength = 500 nm)?
A. 0.01 rads
B. 10-5 rads
C. 10-3 rads
D. 10-4 rads
Physics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

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A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial spe
motikmotik

Answer:

K = G Mm / 9R

Explanation:

Expression for escape velocity V_e = \sqrt{\frac{2GM}{R} }

Kinetic energy at the surface = 1/2 m V_e ²

= 1/2 x m x 2GM/R

GMm/R

Potential energy at the surface

= - GMm/R

Total energy = 0

At height 9R ( 8R from the surface )

potential energy

= - G Mm / 9R

Kinetic energy = K

Total energy will be zero according to law of conservation of mechanical energy

so

K  - G Mm / 9R = 0

K = G Mm / 9R

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3 years ago
17. Saan daw nakasakay ang mga Austronesyan nang dumating sa bansa? A. Bangka B. balangay C galyon D. barko
erastova [34]

Answer:

D.barko po

Explanation:

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7 0
2 years ago
. A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the
telo118 [61]

\text{Given that,}\\\\\text{Mass, m =10 kg}\\\\\text{Time, t = 8 sec}\\\\\text{Velocity, v = 4~m/s}\\\\\text{When a body is moving upwards,}\\\\\text{Tension,}~ T=mg +ma\\\\~~~~~~~~~~~~~~~=mg+m\left(\dfrac{v-u}t \right)\\\\~~~~~~~~~~~~~~~=10(10)+10\left(\dfrac{4-0}8\right)\\\\~~~~~~~~~~~~~~~=100+10\left(\dfrac 12\right)\\\\~~~~~~~~~~~~~~~=100+5\\\\~~~~~~~~~~~~~~~=105~N

5 0
2 years ago
Three 20.0 ohm resistors are
V125BC [204]

Answer:

6.67 ohm

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) =20 ohm

Resistor 2 (R₂) = 20 ohm

Resistor 3 (R₃) = 20 ohm

Equivalent Resistance (R) =?

Since the resistors are arranged in parallel connection, the equivalent resistance can be obtained as follow:

1/R = 1/R₁ + 1/R₂ + 1/R₃

1/R = 1/20 + 1/20 + 1/20

1/R = (1 + 1 + 1) / 20

1/R = 3/20

Invert

R = 20/3

R = 6.67 ohm

Therefore, the equivalent resistance is 6.67 ohm.

5 0
3 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. the car slows uniformly with acceleration of -5.9
Hatshy [7]
Let u =  the speed of the car at the instant when braking begins.

The braking distance is s = 62.3 m, the acceleration is a = -5.9 m/s², and the braking duration is t = 4.15 s.

Use the formula s = ut + (1/2)at² to obtain
(u m/s)*(4.15 s) + 0.5*(-5.9 m/s²)*(4.5 s)² = (62.3 m)
4.15u = 62.3 + 50.8064 = 113.1064
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Let v m/s be the speed with which the car strikes the tree.
Then
v = 27.2546 - 5.9*4.15
   = 2.7696 m/s

Answer: 2.77 m/s (nearest hundredth)

4 0
3 years ago
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