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4 years ago

Arrange the following parts in the order that matches the light path through a light microscope:

Physics

1 answer:

Nonamiya [84]4 years ago

3 0

From the top downward:

Ocular lens (eyepiece)
Body tube
Objective lens
Specimen
Light source

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A metal wire has a resistance of 14.00 Ω at a temperature of 25.0°C. If the same wire has a resistance of 14.55 Ω at 90.0°C, wha

aliina [53]

Answer:

13.52 Ω

Explanation:

coefficient of thermal resistance be α

R₀ , R₂₅ , R₉₀ and R₋₃₂ be resistances at 0 , 25 , 90 , and - 32 degree

R₂₅ = R₀ + α x 25

R₉₀ = R₀ + α x 90

R₉₀ - R₂₅ = 65 x α

α = (R₉₀ - R₂₅ )/ 65

= (14.55 - 14) / 65

= .55 / 65 Ω per °C,

R₂₅ = R₀ + α x 25

14 = R₀ + (.55 / 65 )x 25

= R₀ + .2115

R₀ = 13.7885 Ω

R₋₃₂ = R₀ - α x 32

= 13.7885 -( .55 / 65) x 32

= 13.7885 - .27077

= 13.51773 Ω

= 13.52 Ω

7 0

4 years ago

Read 2 more answers

The curved path traveled by a thrown baseball is known as what?

musickatia [10]

Projectile motion.

hope this helps!

-Payshence

6 0

4 years ago

You are in your car driving on a highway at 23 m/s when you glance in the passenger-side mirror (a convex mirror with radius of

Irina-Kira [14]

Answer:

v = 26. 88 m/s +23 m/s

Explanation:

u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m

$\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}$

$\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}$

Solve s'

$\frac{1}{s'} = \frac{2}{1.50 m} - \frac{1}{2.0 m}$

$\frac{1}{s'} = 1.833 m$

$s' = - 0.545 m$

To determine the speed of the trick to the highway

$\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}$

$\frac{ds}{dt} = 26.88 m/s$

Now to determine the velocity highway is going to be

v = ds/dt + u

v = 26. 88 m/s +23 m/s

8 0

3 years ago

You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

Sophie [7]

The change in potential energy of an object is given by
$U=mg \Delta h$
where
m is the mass of the object
g is the gravitational acceleration
$\delta h$ is the increase in altitude of the object

In our problem, $m=2.5 kg$ is the mass of the book, $g=9.81 m/s^2$ and
$\Delta h=2.6 m -1.2 m=1.4 m$ is the increase in altitude of the book, so its variation of potential energy is
$U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J$

8 0

4 years ago

If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amo

mestny [16]

Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

$P = \frac{V^2}{R}$