Please help!!A camera lens 13.3 cm from a

What are the four forms that water takes when it returns to earth from the clouds?

Marina CMI [18]

Precipitation from the sky to the earth can come in several forms. Four types are: 1) Rain 2) Snow 3) Sleet (think of a more liquidity slushy snow) 4) Hail (pretty much solid chunks of ice)

6 0

3 years ago

You are pushing a 30-kg block on a rough floor in a direction that is parallel to the floor. The block moves with a uniform 2 m/

kykrilka [37]

Answer:dd

Explanation:

I dunno

3 0

3 years ago

Explain how the fixed points are used when calibrating a thermometer.

8090 [49]

<u>Answer:</u>

First, the thermometer is dipped into boiling water, and the mercury inside the thermometer rises to a high level, called the boiling point. This level is then marked as 100°C. The thermometer is then dipped into melting ice, which causes the mercury level to fall to a point called the ice point. This point is then marked as 0°C. The length of the thermometer from the 0°C mark to the 100°C point is then divided into 100 equal sections, and the rest of the levels are marked accordingly.

8 0

2 years ago

A. A light wave moves through glass (n=1.5) at an angle of 25°. What angle will it have when it moves from the glass into air (n

torisob [31]

PART A)

By Snell's law we know that

$n_1sin i = n_2 sin r$

here we know that

$n_1 = 1.5$

$i = 25 degree$

$n_2 = 1$

now from above equation we have

$1.5 sin25 = 1 sin r$

$r = 39.3 degree$

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20}$

$d_i = 30 cm$

so image will form at 30 cm from mirror

it is virtual image and smaller in size

3 0

4 years ago

The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1

xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

C) the flow can be calculated from;

Flow rate ΔV = A•V_avg

A is area = πr²

From question, r = 2cm = 0.02m

A = π x 0.02²

Hence,

ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0

3 years ago