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2 years ago

A tertiary source is the most desirable source for a researcher to use. True or false

1 answer:

6 0

The statement ‘A tertiary source is the most desirable source for a researcher to use’ is false. It should be that your source is primary because it is more reliable and is related to the research you have been working on.

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An aeroplane flew at 600mph for 7 hours. How far did it travel?

____ [38]

Answer:

The aeroplane flew 4200 miles for 7 hours.

600 times 7 is 4200.

6 0

2 years ago

An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the

bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N + N--->N + N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

$L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda$

The frequency is calculated as follows;

$F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz$

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0

2 years ago

8 N to the left , and 4 N to the right. Find the net force. Is this balanced?

alukav5142 [94]

Explanation:

12N by first law of newton is net force after colloision

7 0

1 year ago

A gas always spreads out to fill all available space.<br> True or False <br> Science Hw

GREYUIT [131]

Answer:

true

Explanation:

i know this im in 6th grade

6 0

2 years ago

Read 2 more answers

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

o-na [289]

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$

There is NO initial vertical velocity, so:

$\large\boxed{\Delta d= \frac{1}{2}at^2}}$

Rearrange to solve for time:

$2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}$

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

$t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s$

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

$\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}$

6 0

2 years ago