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2 years ago

You are measuring the volume of a chemical beaker how would u take the measurment?

1 answer:

6 0

I think its d because lifting it would make the chemical swish around and that will make it so you cant get the right measurement. hope this helps :)

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Sort the processes based on the type of energy transfer they involve.??

klemol [59]

The 1st one goes two added sodoes the second one then the third goes to removed then the fourth goes to added and the rest go to removed

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3 years ago

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umka2103 [35]

You can hear a difference between these two sounds. That is because their pitch isdifferent. Pitch depends on the frequency of a sound wave. ... High sounds have highfrequencies and low sounds have lowfrequencies.

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3 years ago

A crane lifts a 545 kg piano up into a high-rise apartment 75.0 meters above the ground. In order to do so, 5350 Newtons were ap

vodka [1.7K]

Answer:

Work done to pull the piano upwards is 401250 J

Explanation:

Work is done against the gravity to pull the piano upwards

So here we can say that work done is

$W = mgH$

here we know that

$mg = 5350 N$

also we know that

H = 75 m

now we have

$W = 5350 \times 75$

$W = 401250 J$

7 0

3 years ago

A certain wave has a wavelength of 45.0 meters and a frequency of 9.00 Hz. What is the speed of the wave? 405 m/s 5.00 m/s 0.200

kirill115 [55]

Wave speed = (wavelength) x (frequency)

= (45 meters) x (9 per second)

= 405 meters per second .

5 0

2 years ago

Read 2 more answers

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

2 years ago