Which part of the microscope is responsible for enlarging the specimen?

What kind of light source is an IMAX movie screen

kirill115 [55]

Answer:

laser light source

Explanation:

A laser light source provides substantially more brightness than a xenon bulb, and allows IMAX to power its largest screens with even sharper and more lifelike images.

4 0

3 years ago

Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str

Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, $v_y$ , of the package just before landing is given by the relation;

$v_y$ ² = u² + 2·g·h

u = 0 m/s

∴ $v_y$ ² = 0 + 2×9.81×1500 = 29430 m²/s²

$v_y$ = √29430 = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

$tan \theta = \dfrac{v_y}{v_x} = \dfrac{171.55}{39.0 } = 4.4$

∴ θ = tan⁻¹(4.4) = 77.19°.

6 0

4 years ago

Three 100 nCcharged objects are equally spaced on a straight line. The separation of each object from its neighbor is 0.3 m. Fin

Rainbow [258]

Answer:

0N

Explanation:

Parameters given:

Q1 = Q2 = Q3 = 100 * 10^(-9) C

r = 0.3 m

The net force on the middle charge is:

F = F(1,2) + F(2,3)

Where F(1,2) = force due to Q1 on Q2

F(2,3) = force due to Q3 on Q2

F(1,2) = k*Q(1)*Q(2)/r²

F(1,2) = [9 * 10^9 * 100 * 10^(-9) * 100 * 10^(-9)]/(0.3²)

F(1,2) = 0.001N

F(2,3) = k*Q(2)*Q(3)/r²

F(2,3) = [9 * 10^9 * 100 * 10^(-9) * -100 * 10^(-9)]/(0.3²)

F(2,3) = -0.001N

Therefore,

F = 0.001 - 0.001

F = 0N

6 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

so

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago

The mass of a newly discovered planet is 4.4 X 1025 kg. What must the radius of the planet be to have

777dan777 [17]

.67 because this is right

8 0

3 years ago