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3 years ago

Objective lenses are contained in a ____________ that can be turned to put a particular objective lens in place to be used.

Physics

1 answer:

Masteriza [31]3 years ago

6 0

Answer:

Revolving nosepiece

Explanation:

The revolving nosepiece is one of the parts of a microscope, used for holding the objective lenses. They can be turned to put a particular objective lens in place to be used in order to vary magnification.

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You push downward on a trunk at an angle 25° below the horizontal with a force of If the trunk is on a flat surface and the coef

Sophie [7]

Complete question is;

You push downward on a trunk at an angle 25° below the horizontal with a force of 750N. if the trunk is on a flat surface and the coefficient of static friction between the surface and the trunk is 0.61, what is the most massive trunk you will be able to move?

Answer:

The most massive trunk is about 81.3 kg

Explanation:

I've attached a free body diagram that depicts this question.

Where;

N = normal force on the trunk

m = mass of the trunk

W = weight of the trunk = mg

F = static frictional force

Using equilibrium of force in vertical direction, we obtain;

N = W + 750 Sin25

N = mg + 750 Sin25 - - - - (eq 1)

Now, we are given that Coefficient of static friction: μ = 0.61

static frictional force is given by the formula;

F = μN

Since N = mg + 750 Sin25, we now have;

F = (0.61) (mg + 750 Sin25) - - - (eq 2)

Along the horizontal direction, for the trunk to move, force equation must be;

F = 750 Cos25

Thus, we now have;

750 Cos25 = 0.61(mg + 750 Sin25)

g = 9.81.

So,we now have ;

750 Cos25 = 0.61(m(9.81) + 750Sin25)

750 × 0.9063 = 0.61(9.81m + (750 × 0.4226))

Divide both sides by 0.61;

(750 × 0.9063)/0.61 = 9.81m + 316.95

1114.3 = 9.81m + 316.95

1114.3 - 316.95 = 9.81m

797.35 = 9.81m

m = 797.35/9.81

m = 81.3 kg

7 0

3 years ago

1 2 3 What would happen if Bulb 1 goes out?

aleksandrvk [35]

Answer:

it will explode

Explanation:

because it will explode

5 0

3 years ago

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadi

zepelin [54]

Answer:

his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.

determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.

a)0.7956kg/s

b)5.437 × 10⁻³m²

Explanation:

The concepts related to the change of mass flow for both entry and exit is applied

The general formula is defined by

$\dot{m}=\rho A V$

Where,

$\dot{m} = mass flow rate\\\rho = Density\\V = Velocity$

values are divided by inlet(1) and outlet(2) by

$\rho_1 = 2.21kg/m^3V_1 = 40m/s$

$A_1 = 90*10^{-4}m^2\\\rho_2 = 0.762kg/m^3\\V_2 = 192m/s$

PART A) Applying the flow equation

$\dot{m} = \rho_1 A_1 V_1\\\dot{m} = (2.21)(90*10^{-4})(40)\\\dot{m} = 0.7956kg/s$

PART B) For the exit area we need to arrange the equation in function of Area, that is

$A_2 = \frac{\dot{m}}{\rho_2 V_2}\\A_2 = \frac{0.7956}{(0.762)(192)}\\A_2 = 5.437*10^{-3}m^2$

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3 years ago

An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic sound that travels 1530 m/s in seawater. find the dept

kompoz [17]

It will travel 1530 m* 6 seconds =9180m. thus the depth there will be 9180/2=4590m.

3 0

4 years ago

The outermost portion of the kidney is know as the

Ivahew [28]

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