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2 years ago

8

If two stars in the sky are separated by twice the angle projected by your index and pinky fingers, then how many degrees are th

ey apart?

1 answer:

quester [9]2 years ago

7 0

Answer:

The answer is 30 degrees.

Explanation:

The angle between the index and pinky fingers is 15 degrees.

Twice this angle is 2×15=30degrees.

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sest yourse 1. A pencil lies on the dashboard of a car. a) i) What happens to the pencil when the car suddenly stops? suddenly a

julia-pushkina [17]

Answer:

1. the pencil would have the momentum and would keep going until it hits the windshield. 2. when the car suddenly accelerates, the pencil would be inert and it would move toward the back of the car until a constant speed from the car is reached.

8 0

2 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

2 years ago

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current

balu736 [363]

Answer:

(a) The resistance R of the inductor is 2480.62 Ω

(b) The inductance L of the inductor is 1.67 H

Explanation:

Given;

emf of the battery, V = 16.0 V

current at 0.940 ms = 4.86 mA

after a long time, the current becomes 6.45 mA = maximum current

Part (a) The resistance R of the inductor

$R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms$

Part (b) the inductance L of the inductor

$\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}$

where;

L is the inductance

R is the resistance of the inductor

t is time

$L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H$

Therefore, the inductance is 1.67 H

5 0

3 years ago

They realize there is a thin film of oil on the surface of the puddle. If the index of refraction of the oil is 1.81, and they o

Sphinxa [80]

Answer:

The right solution is "165.8 nm".

Explanation:

Given:

Index of refraction,

n = 1.81

Wavelength,

λ = 600 nm

We know that,

⇒ $t=\frac{\lambda}{2\times n}$

By putting the values, we get

$=\frac{600}{2\times 1.81}$

$=165.8 \ nm$

3 0

2 years ago

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from th

vladimir1956 [14]

Complete question:

A light bulb emits light that travels uniformly in all directions. Detailed measurements show that at a distance of 56 m from the bulb, the amplitude of the electric field is 3.78 V/m. What is the average intensity of the light?

Answer:

The average intensity of the light is 0.02 W/m²

Explanation:

Given;

Amplitude of the electric field, E₀ = 3.78 V/m

The average intensity of the light is calculated as follows;

$I_{avg} = \frac{c\epsilon_0 E_0^2}{2}$

where;

$I_{avg}$ is the average intensity of the light

c is speed of light = 3 x 10⁸ m/s

$I_{avg} = \frac{(3\times 10^8)(8.85 \times 10^{-12}) (3.78)^2}{2} \\\\I_{avg} = 0.01897 \ W/m^2\\\\I_{avg} = 0.02 \ W/m^2$

Therefore, the average intensity of the light is 0.02 W/m²

4 0

2 years ago