What is the definition of electrical power?

you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

Sergio039 [100]

Some people view bacteria specimens with a 100x objective lens in order to see the smallest details.
Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
And still others may use a 40x objective lens to gain maximum resolution when viewing images of thick samples.

It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>

Bacteria must, of course, be viewed at the maximum magnification and resolution possible because to their small size.
Due to optical restrictions, this is approximately 1000x in a light microscope.
To improve resolution, the oil immersion method is performed. This calls for a unique 100x objective.

To learn more about bacterial specimen, visit:

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5 0

2 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

As an object sinks in a fluid, the buoyant force ____.

kiruha [24]

Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.

As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.

3 0

3 years ago

Read 2 more answers

The planet Krypton has a mass of 8.8 × 1023 kg and radius of 2.5 × 106 m. What is the acceleration of an object in free fall nea

djverab [1.8K]

Answer:

Acceleration, $a=9.39\ m/s^2$

Explanation:

Given that,

Mass of the planet Krypton, $m=8.8\times 10^{23}\ kg$

Radius of the planet Krypton, $r=2.5\times 10^{6}\ m$

Value of gravitational constant, $G=6.6726\times 10^{-11}\ Nm^2/kg^2$

To find,

The acceleration of an object in free fall near the surface of Krypton.

Solution,

The relation for the acceleration of the object is given by the below formula as :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.6726\times 10^{-11}\times 8.8\times 10^{23}}{(2.5\times 10^{6})^2}$

$a=9.39\ m/s^2$

So, the value of acceleration of an object in free fall near the surface of Krypton is $9.39\ m/s^2$

5 0

4 years ago

The element hydrogen has the highest specific heat of all elements. At a temperature of 25°C, hydrogen’s specific heat capacity

natali 33 [55]

Answer:

121550 J

Explanation:

Parameters given:

Mass, m = 0.34kg

Specific heat capacity, c = 14300 J/kgK

Change in temperature, ΔT = 25K

Heat gained/lost by an object is given as:

Q = mcΔT

Since ΔT is positive in this case and also because we're told that heat was transferred to the hydrogen sample, the hydrogen sample gained heat. Therefore, Q:

Q = 0.34 * 14300 * 25

Q = 121550J or 121.55 kJ

8 0

3 years ago