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3 years ago

What are the magnifications of the objective lenses

Physics

1 answer:

Anvisha [2.4K]3 years ago

7 0

Answer:

Objective Lenses: Usually you find 3 or 4 objective lenses on a microscope. They almost always consist of 4x, 10x, 40x and 100x powers, when coupled with a 10x (most common) eyepiece lens, total magnification is 40x (4x times 10x) 100x, 400x and 1000x

Explanation:

Objective lenses come in various magnification powers, with the most common being 4x, 10x, 40x, and 100x, also known as scanning, low power, high power and typically oil immersion objectives, respectively

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In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha

trapecia [35]

Answer:

Velocity will be $v=4.06\times10^7m/sec$

Explanation:

We have given mass of alpha particle m = 4 u $=4\times 1.67\times 10^{-27}kg=6.68\times 10^{-27}kg$

Charge on alpha particle $q=2e=2\times 1.6\times 10^{-19}C=3.2\times 10^{-19}C$

Charge on thorium particle $=90\times 1.6\times 10^{-19}=144\times 10^{-19}C$

Diameter is given as d = 15 fm

So radius $r=\frac{d}{2}=\frac{15}{2}=7.5fm=7.5\times 10^{-15}m$

Potential energy is given by $E=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r}=\frac{Kq_1q_2}{r}=\frac{9\times 10^9\times 144\times 10^{-19}\times 3.2\times 10^{-19}}{7.5\times 10^{-15}}=5.529\times 10^{-12}J$

From energy conservation $\frac{1}{2}mv^2=5.529\times 10^{-12}$

$\frac{1}{2}\times 6.68\times 10^{-27}v^2=5.529\times 10^{-12}$

$v=4.06\times10^7m/sec$

6 0

3 years ago

A 1.94-m-diameter lead sphere has a mass of 5681 kg. A dust particle rests on the surface. What is the ratio of the gravitationa

scoundrel [369]

To develop this problem we will apply Newton's laws regarding gravitational forces, both in space and on earth. From finding this relationship, leaving the variable of the dust mass open, we will find the relationship of the forces between the two surfaces. Our values are,

$\text{Diameter of the lead sphere} = D=1.940m$

$\text{Mass of the lead sphere} = m_1 = 5681kg$

$\text{Mass of the dust particle} = m_2$

Distance between the center of lead sphere to dust particle

$r = \frac{D}{2}$

$r = \frac{1.940m}{2}$

$r = 0.97m$

Gravitational force of the sphere on the dust particle:

$F = \frac{Gm_1m_2}{r^2}$

$F = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5681kg)(m_2)}{(0.97m)^2}$

$F = (4.027*10^{-7} N/kg)m_2$

Weight of the dust particle

$W = m_2 g$

$W = m_2 (9.8m/s^2)$

Ratio of F and W:

$\frac{F}{W} = \frac{(4.07*10^{-7}N/kg) m_2)}{m_2(9.8m/s^2)}$

$\frac{F}{W} = 4.153*10^{-8}$

Therefore the ratio is $4.153*10^{-8}$

3 0

3 years ago

A camcorder has a power rating of 10 watts. If the output voltage from its battery is 3 volts, what current does it use?

Gemiola [76]

The power is calculated using the following rule:
Power = voltage * current
We are given that:
power = 10 watts
voltage = 3 volts

Substitute with the givens in the above equation to get the value of the current as follows:
P = V*I
10 = 3*I
current (I) = 10/3 amperes

7 0

3 years ago

Radiation heat transfer can be due to what?

riadik2000 [5.3K]

Answer: Heat transfer due to emission of electromagnetic waves is known as thermal radiation.

Explanation: Heat transfer through radiation takes place in form of electromagnetic waves mainly in the infrared region. Radiation emitted by a body is a consequence of thermal agitation of its composing molecules.

4 0

3 years ago

Read 2 more answers

A sample of polonium-210 was left for 414 days.

jenyasd209 [6]

Answer:

PO(1) = PO(0) / 2 1 refers to 1 half life of PO(0)

PO(2) = PO(1) / 2 = P(0) / 4 amount of PO left after 2 half-lives

PO(3) = PO(2) / 2 = PO(0) / 8 amount of PO left after 3 half-lives

414 da / 138 da = 3 3 half-lives pass in 414 da

PO(0) = 8 PO(3) = 8 * 1.45E-4 g = 1.16E-3 g = .00116 g after 414 days

6 0

2 years ago