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3 years ago

How do you find the total magnification of a microscope?

1 answer:

6 0

In order to find total magnification of a microscope, you need to multiply the power of eyepiece and objective lens.

Hope this helps!

You might be interested in

An object is given a very small amount of charge. Which of the following could

spayn [35]

5.4*10^-19 C

Explanation:

For the purposes of this question, charges essentially come in packages that are the size of an electron (or proton since they have the same magnitude of charge). The charge on an electron is -1.6*10^-19

Therefore, any object should have a charge that is a multiple of the charge of an electron - It would not make sense to have a charge equivalent to 1.5 electrons since you can't exactly split the electron in half. So the charge of any integer number of electrons can be transferred to another object.

Charge = q(electron)*n(#electrons)

Since 5.4/1.6 = 3.375, we know that it can not be the right answer because the answer is not an integer.

If you divide every other option listed by the charge of an electron, you will get an integer number.

(16*10^-19 C)/(1.6*10^-19C) = 10

(-6.4*10^-19 C)/(1.6*10^-19C) = -4

(4.8*10^-19 C)/(1.6*10^-19C) = 3

(5.4*10^-19 C)/(1.6*10^-19C) = 3.375

(3.2*10^-19C)/(1.6*10^-19C) = 2

etc.

I hope this helps!

3 0

3 years ago

How to measure the external diameter of a sphere

DanielleElmas [232]

Sphere is that the circular objects in the two dimensional space (1) circle
(2) disk. Two dimensional space is a set of points and the distance of that point,The two points of Sphere that length and center.
Sphere can constructed as the named of surface form circle about any diameter. circle is the special type of the revolution replacing the circle,
sphere is the distance r is the radius of the ball and circle is the center of mathematical ball,as the center and the radius of the sphere is to respectively.
The ball and sphere has not be maintained mathematical references as a solid references. A sphere of any radius is centered at the number of zero.

4 0

3 years ago

The weights in atwoods machine, starting at rest, attain a velocity of 2ft/sec in one sec. Find the ratio of the masses

Orlov [11]

Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.

Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)

Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a

Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a

Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)

With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962

Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).

8 0

3 years ago

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the sa

tester [92]

Answer:

Part a)

$t = 2.83 s$

Part b)

Ball thrown downwards = $v_f = 23.8 m/s$

Ball thrown upwards = $v_f = 23.8 m/s$

Part c)

$d = 22.24 m$

Explanation:

Part a)

Since both the balls are projected with same speed in opposite directions

So here the time difference is the time for which the ball projected upward will move up and come back at the same point of projection

Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

$\Delta y = 0 = v_y t + \frac{1}{2}at^2$

$0 = 13.9 t - \frac{1}{2}(9.81) t^2$

$t = 2.83 s$

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

$v_f^2 - v_i^2 = 2 a \Delta y$