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Aloiza [94]
3 years ago
9

. A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The f

ocal length of the eyepiece is 5 cm. Calculate the magnitude of the angular magnification of the telescope.
Physics
1 answer:
beks73 [17]3 years ago
5 0

Answer:

the magnitude of the angular magnification of the telescope. is 4

Explanation:

Calculate the magnitude of the angular magnification of the telescope.

Given that,

distance = 25cm

focal length from the objective f₀ = 20cm

focal length from eye piece f₁ = 5cm

The angular magnification of the telescope is

M = \frac{f_0}{f_1}

Magnification = 20 / 5

magnification = 4

Hence, the magnitude of the angular magnification of the telescope. is 4

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Hi There! :)

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6 0
3 years ago
Calculate the amount of air in a room 6m long, 5m wide and 3mm high.​
sattari [20]

Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

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2 years ago
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Un lote de construcción rectangular mide 100 metros por 150 metros. Calcula el área de este lote en yardas cuadradas.
antoniya [11.8K]

Answer: 17939.74 yards

Explanation:

Given , A rectangular measures 100 meters by 150 meters

To find : Area of rectangle.

Formula :

Area of rectangle = Length x width

Here, let length = 100 meters and width = 150 meters

Then, Area of rectangle = 100 meters x 150 meters = 15,000 square meters

Also , 1 meter = 1.09361 yards

Then, Area of rectangle = 15,000 x 1.09361 x 1.09361 square yards

= 17939.7424815 square yards

≈ 17939.74 yards

Hence, the area of rectangle is 17939.74 yards .

8 0
3 years ago
Which image shows an example of the strong nuclear force in action?
dusya [7]

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6 0
3 years ago
I need answers and solvings to these questions​
den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) 15.7 rad/s^2

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular velocity

\omega_i is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

\omega_i = 0 (since it starts from rest)

\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s is the final angular velocity

t = 2 s

Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

a=-\omega^2 x

where

a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

f=\frac{\omega}{2\pi}

Therefore, the frequency of this simple harmonic oscillator is

f=\frac{10}{2\pi}=1.6 Hz

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5

The angular frequency of the spring is given by

\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz

And therefore, its period is

T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

v_1 = 4 m/s is the speed of the petrol in the hose

A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

Solving for r_2, we find the radius of the nozzle:

\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm

So, the diameter of the nozzle will be

d_2 = 2r_2 = 2(2.5)=5.0 mm

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

\frac{F_1}{A_1}=\frac{F_2}{A_2}

where

F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N is the force applied on the bigger piston (the weight of the car)

A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

Solving for F_1, we find

F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0
3 years ago
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