All categories

3 years ago

Camera lenses are described in terms of

Physics

1 answer:

Lesechka [4]3 years ago

6 0

Answer:

The position of the image from the camera lens is approximately 121.1 mm

Explanation:

The given parameters of the lens are;

The specification of the camera lens = 120 mm

Therefore, the focal length of the camera lens, f = 120 mm = 0.12 m

The distance of the object from the camera, $d_o$ = 13 m

The lens equation for finding the position of the image is given as follows;

$\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$

Where;

$d_i$ = The position of the image from the camera lens

Therefore, by plugging in the known values, we have;

$\dfrac{1}{13} = \dfrac{1}{0.12} + \dfrac{1}{d_i}$

$\dfrac{1}{d_i} = \dfrac{1}{0.12} - \dfrac{1}{13} = \dfrac{13 - 0.12}{0.12 \times 13} = \dfrac{12.88}{1.56} = \dfrac{322}{39}$

$\therefore d_i = \dfrac{39}{322} \approx 0.1211$

The position of the image from the camera lens, $d_i$ = 0.1211 m = 121.1 mm

You might be interested in

If an athlete executing a long jump leaves the ground at a 30° angle and travels 7.80 m, what was the takeoff speed

Mumz [18]

R=u^2 sin2x / g
7.8=u^2 sin2*30 /10
u=9.5m/s

6 0

3 years ago

Describe the core-mantle-crust structures of the terrestrial worlds. What is differentiation? What do we mean by the lithosphere

Serhud [2]

Answer:

The core: this structure of the terrestrial world is made up of sulphur, iron, silicon and it is found at the centre of the terrestrial world.

The mantle: this structure of the terrestrial world is located between the outer core and the crust and consist of mainly silicate rock.

The crust: this is the structure of the terrestrial world that makes up the outermost layer. It consist of granite and basalt.

Differentiation: this is the process by which gravity separates materials according to density, with high-density materials sinking and low-density materials rising.

Lithosphere: this is made up of the hard and rigid outer layer of the terrestrial planet and it's located between the crust and the outermost mantel.

The five terrestrial worlds includes Mercury, Venus, Earth, the Moon, and Mars. The largest which includes the Venus and Earth has the thinnest lithosphere while the smallest ( mercury and moon) have the thickest lithosphere.

8 0

4 years ago

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

4 years ago

6) A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10−3 s

Romashka [77]

Answer:

a) L = 2.10x10⁴⁰ kg*m²/s

b) τ = 1.12x10²⁴ N.m

Explanation:

a) The angular momentum (L) of the pulsar can be calculated using the following equation:

$L = I \omega$

Where:

I: inertia momentum

ω: angular velocity

First we need to calculate ω and I. The angular velocity can be calculated as follows:

$\omega = \frac{2 \pi}{T}$

Where:

T: is the period = 33.5x10⁻³ s

$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s$

The inertia moment of the pulsar can be calculated using the following relation:

$I = \frac{2}{5}mr^{2}$

Where:

m: is the mass of the pulsar = 2.8x10³⁰ kg

r: is the radius = 10.0 km

$I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2}$

Now, the angular momentum of the pulsar is:

$L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1}$

b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:

$\tau = I*\alpha$

Where:

α: is the angular acceleration = 10⁻¹⁴ rad/s²

$\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m$

I hope it helps you!

5 0

3 years ago

A car has uniform velocity of 108km/hour. How far does it travel in 1 1/2 minutes

Nostrana [21]

Answer:

2.7km

Explanation:

Two methods: Convert km/hour to km/minutes or convert 3/2 minutes to hours.

Then multiply time to get the distance of the car traveled.

3 0

3 years ago