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3 years ago

How does the angular diameter of an object change as you move away from the object?

Physics

2 answers:

GarryVolchara [31]3 years ago

6 0

Answer:

its B

Explanation:

7nadin3 [17]3 years ago

5 0

Explanation :

Angular diameter of a sphere or a circle shows how large a sphere or a circle appears from a given point. It is the angle between the initial and final positions.

The angular diameter of a given sphere or a circle is given by :

$\delta =2arc\ tan\dfrac{d}{2D}$

where $\delta$ is angular displacement

d is actual diameter of object

D is distance to the center of sphere

So, angular displacement is inversely proportional to the distance from the object.

<em>Hence, as we move away from the object the angular diameter of an object gets smaller.</em>

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Answer:

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Explanation:

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If the Moon had twice as much mass and still orbits Earth at the same distance, ocean bulges on Earth would be

Sophie [7]

Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.

<h3>What is ocean bludge?</h3>

The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.

The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.

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6 0

1 year ago

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.

Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA\dfrac{dB}{dt}$

For N = 1

$\epsilon=A\dfrac{dB}{dt}$

We need to calculate the current

Using formula of current

$i=\dfrac{\epsilon}{R}$

Put the value of emf

$i=\dfrac{A\dfrac{dB}{dt}}{R}$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA\dfrac{dB}{dt}$

Then the current would be

$i'=\dfrac{\epsilon'}{NR}$

$i'=\dfrac{NA\dfrac{dB}{dt}}{NR}$

$i'=\dfrac{A\dfrac{dB}{dt}}{R}$

$i'=i$

Hence, The current would be same in both situation.

4 0

3 years ago

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle

Alex73 [517]

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

$T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N$

6 0

3 years ago