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Alex_Xolod [135]
3 years ago
13

Any object that just fits into your visual field has this angular size. Do all objects that you see this this angular size have

the same linear size (the size you would measure by placing a ruler next to them)? If not, when does an object have a larger linear size than another object even though both have the same angular size?
Physics
1 answer:
Nataly_w [17]3 years ago
3 0

Answer:

No the angular and linear size will not be same for different objects bearing the same angular size and is dependent on the point of observation. The greater the distance between the observer and the object, the greater the difference between the angular and linear size.

Explanation:

The angular size is the ratio of two lengths namely

Angular \, Size=\frac{Linear Size}{Distance from the object}

Here if the distance from the object is unity, the angular and linear size will be similar. However if the distance of observation is very large, the angular size for large bodies with large linear size will be very small.

An example in this regard is Sun.

The linear size of Sun, (the diameter) is 1.3927 million km. Which is very large. However as it is very far from earth, 147.44 million km, the angular size is very small.

it is given as

Angular \, Size=\frac{1.39}{147.44}=0.009 rad

Now the same angular size can be of a tennis ball having a diameter of 10 cm , placed at around 10.6 m away.

Angular \, Size=\frac{0.1}{10.6}=0.009 rad

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A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge
Serga [27]

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

6 0
3 years ago
A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec
olga2289 [7]

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

h=u't+\dfrac{1}{2}gt^2

h=\dfrac{1}{2}gt^2

h=\dfrac{1}{2}\times 9.8\times (4.3)^2

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

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What is the mass of a football in free fall with a total force due to gravity of 782N?
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4 0
3 years ago
Read 2 more answers
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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