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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 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in
chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

 m v + Mu = m v' + M u'

 0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) determine the distance traveled during the first second.
barxatty [35]

The relevant formula we can use in this case would be:

h = v0 t + 0.5 g t^2

where,

h = height or distance travelled

v0 = initial velocity = 0 since it was dropped

t = time = 1 seconds

g = 9.8 m/s^2

 

So calculating for height h:

h = 0 + 0.5 * 9.8 m/s^2 * (1 s)^2

<span>h = 4.9 meters</span>

6 0
3 years ago
5-A baseball is hit with a velocity of 15 m/s at an angle of 40° in the horizontal direction. What are the horizontal and vertic
borishaifa [10]

<u>Answer</u>

B•Horizontal=11.49 m/s  

Vertical=9.64 m/s

Using the concept of a trigonometric ratios,

sin θ  = y/hypotenuse

where y is the vertical component.

sin 40 = y/15

y = 15 × sin 40

  = 9.64 m/s

vertical component = 9.64 m/s

cos θ  = x/hypotenuse

where x is the horizontal component

cos 40 = x/15

x = 15 × cos 15

    =11.49

Horizontal component = 11.49 m/s

5 0
3 years ago
After an initial test run John determines that his cooling system generates 45 W of heat loss. Calculate the amount of heat loss
anyanavicka [17]

Given:

heat generated by John's cooling system,  H = \rho A v^{3}  = 45 W    (1)

If ρ, A, and v corresponds to John's cooling system then let \rho_{1}, A_{1}, v_{1} be the variables for Mike's system then:

\rho  = 9.5\rho_{1}

\rho_{1}  = \frac{\rho}{9.5}

v_{1} =3.5 v

Formula use:

Heat generated, H = \rho A v^{3}

where,

\rho = density

A = area

v = velocity

Solution:

for Mike's cooling system:

H_{2} = v_{1}^{3}{1}A_{1}\rho_{1}

⇒ H_{2} = (3.5v)^{3} × A × \frac{\rho}{9.5}

H_{2} = 4.513v^{3} A  \rho

Using eqn (1) in the above eqn, we get:

H_{2} = 4.513 × 45 = 203.09 W

5 0
3 years ago
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