Ask question

Ask question

All categories

3 years ago

7

An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t

wo significant figures and include the appropriate units.

1 answer:

elixir [45]3 years ago

5 0

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

$\mu = \frac{f_0}{f_e}$

Here,

$\mu$ = Magnification

$f_e$ = Focal length eyepieces

$f_0$ = Focal length of the Objective

Rearranging to find the focal length of the objective

$f_0 = \mu f_e$

Replacing with our values

$f_0 = 7.5* 3.7cm$

$f_0 = 27.75cm$

Therefore the focal length of th eobjective lenses is 27.75cm

You might be interested in

Two concrete spans of a 380 m long bridge are

Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

$L_i = \frac{L}{2}=\frac{380}{2}=190 m$

Due to the increase in temperature, the length of each span increases according to:

$L_f = L_i(1+ \alpha \Delta T)$

where

$L_i = 190 m$ is the initial length of one span

$\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1}$ is the temperature coefficient of thermal expansion

$\Delta T=20^{\circ}C$ is the increase in temperature

Substituting,

$L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m$

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

$h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m$

4 0

3 years ago

Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro

alex41 [277]

They were formed in the nuclear<span> fusion reaction inside older </span><span>stars.

As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.</span>

8 0

3 years ago

The work-energy theorem states that a force acting on a particle as it moves over a ______ changes the ______ energy of the part

svetoff [14.1K]

Answer:

B. distance/potential

Explanation:

Quizlet

5 0

3 years ago

One molecule of bromine (Br2) and two molecules of potassium chloride (KCI) combine in a reaction. How many atoms are in the pro

Liula [17]

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Let's consider the balanced equation for the reaction between 1 molecule of bromine and 2 molecules of potassium chloride. This is a single replacement reaction.

Br₂ + 2 KCl ⇒ 2 KBr + Cl₂

We obtain as products, 2 molecules of potassium bromide and 1 molecule of chlorine.

1 molecule of KBr has 2 atoms, so 2 molecules contribute with 4 atoms.
1 molecule of Cl₂ has 2 atoms.
The 4 atoms from KBr and the 2 atoms from Cl₂ make a total of 6 atoms.

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Learn more: brainly.com/question/21850455

5 0

3 years ago

Read 2 more answers

On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini

irinina [24]

Answer:

$v_a=0.8176 m/s$

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being $m_a$ and $m_b$ the masses of pucks a and b respectively, the initial momentum of the system is

$M_1=m_av_a+m_bv_b$

Since b is initially at rest

$M_1=m_av_a$

After the collision and being $v'_a$ and $v'_b$ the respective velocities, the total momentum is

$M_2=m_av'_a+m_bv'_b$

Both momentums are equal, thus

$m_av_a=m_av'_a+m_bv'_b$

Solving for $v_a$

$v_a=\frac{m_av'_a+m_bv'_b}{m_a}$

$v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}$

$v_a=0.8176 m/s$

The initial kinetic energy can be found as (provided puck b is at rest)

$K_1=\frac{1}{2}m_av_a^2$

$K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J$

The final kinetic energy is

$K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2$

$K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J$

The change of kinetic energy is

$\Delta K=0.07969 J - 0.0849 J = -0.00521 J$

3 0

3 years ago