Ask question

Ask question

All categories

3 years ago

9

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

ngth is 25 cm. What is the magnitude of the overall magnification of the microscope?

1 answer:

expeople1 [14]3 years ago

6 0

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

So

$m = 25 * 12.5$

$m \approx 310$

You might be interested in

A wrestler weighs in for the first match on the moon. will the athlete weigh more or less on the moon than he does on Earth?

Firdavs [7]

The moon<span> is 1/4 the size of </span>Earth<span>, so the </span>moon's<span> gravity is much less than the </span>earth's gravity, 83.3% (or 5/6) less to be exact. Finally, "weight<span>" is a measure of the gravitational pull between two objects. So of course you would </span>weigh<span> much less on the </span>moon<span>.</span>

4 0

3 years ago

You exert a force of 75 newtons on a rock. You push and you push, but you can’t budge it. You are exhausted! How much work did y

dalvyx [7]

Answer: Work W = 0

Explanation: Work W = F·s. Because rock does not move, s = 0 and

work done is zero.

4 0

3 years ago

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

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

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

8 0

3 years ago

The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass

Triss [41]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

spring constant of the spring attached to the input piston, $k=1600\ N.m^{-1}$

mass subjected to the output plunger, $m=40\ kg$

<u>Now, the force due to the mass:</u>

$F=m.g$

$F=40\times 9.8$

$F=392\ N$

<u>Compression in Spring:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{392}{1600}$

$\Delta x=0.245\ m$

or

$\Delta x=245\ mm$

8 0

4 years ago

A capacitor consists of two square plates, 8.7 cm on a side, separated by a 2.0 mm air gap. How much energy would be stored in t

slavikrds [6]

Answer:

122.84 J

Explanation:

Since plate is square, area, A is given by $(8.7/100)^{2}=0.007569m^{2}$

The distance between plates, d, is given in the question as 2mm=0.002m

Charge on plate, Q, as given in the question is 240 $\mu c$

Assuming mica dielectric constant, k of 7

Capacitance, C is given by

C= $\frac {k\epsilon_{o}A}{d}=\frac {(7)(8.85*10^{-12})(0.007569)}{0.002}=2.34*10*^{-10}F$

Stored energy, E is given by

E= $\frac {Q^{2}}{2C}=\frac {(240*10^{-6})^{2}}{2*(2.34*10^{-10})}=122.84J$

Therefore, the stored energy is 122.84 J

5 0

3 years ago