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3 years ago

10

The eyepiece of a light microscope has a magnification level of 10x. If you were looking at a paramecium under the lowest-power

objective (4x), what would be the total magnification of the microscope at this setting?

1 answer:

tester [92]3 years ago

4 0

Answer:

The total magnification will be 40x.

Explanation:

As we know if the magnification of the eyepiece of a microscope be $m_{e}$ and that of the objective be $m_{o}$ , then the total magnification is given by

$m = m_{e} \times m_{o}$

Given, $m_{e} = 10x$ and $m_{o} = 4x$

Therefore total magnification is $m = 10 \times 4 = 40x$

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Cathode rays were shown to be a stream of _____.

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Cathode rays were shown to be a stream of "electrons".

Cathode rays (likewise called an electron beam) are streams of electrons saw in vacuum tubes. In the event that a cleared glass tube is outfitted with two anodes and a voltage is connected, the glass inverse the negative terminal is seen to sparkle from electrons radiated from the cathode. Electrons were first found as the constituents of cathode beams. The picture in an exemplary TV is made by centered light emission redirected by electric or magnetic fields in cathode ray tubes (CRTs).

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4 years ago

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The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

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3 years ago

0. The temperature of source is 500K with source energy 2003, what is the temperature of sink with sink energy 100 J? a. 500 K b

andreev551 [17]

Answer:

c. 250k

Explanation:

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3 years ago

The mass of the basketball shown below is 4 times the mass of the baseball.

igomit [66]

Answer:

<h3>40.0 Joules. None of the answers is correct</h3>

Explanation:

Kinetic energy of an object is expressed as;

KE = 1/2mv² where;

m is the mass of the object

V is the velocity of the object.

Let Ma be the mass of the basketball and Mb be the mass of baseball.

If the mass of the basketball shown below is 4 times the mass of the baseball, then Ma = 4Mb

KE of the basket ball = 1/2MaVa²

Va² = 2(KE)a/Ma ...... 1

KE of the baseball = 1/2MbVb²

Vb² = 2(KE)b/Mb .......... 2

Since their velocities are the same, hence Va²= Vb²

2(KE)a/Ma = 2(KE)b/Mb

Substituting Ma = 4Mb into the resulting expression.

2(KE)a/4Mb = 2(KE)b/Mb

2(KE)a/4 = 2(KE)b

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2(KE)a/4 = 2(10)

2(KE)a = 4*20

2(KE)a = 80

(KE)a = 80/2

(KE)a = 40.0J

Hence the kinetic energy of the basketball is 40.0Joules

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