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3 years ago

15

If you are going to see which food dye diffuses fastest in water. Which variables do you measure How will you do this?

1 answer:

mariarad [96]3 years ago

6 0

You should be searching for your independent variable, and think about every variable to keep the same. The only thing that should be changing between separate experiments would be the type of food dye, you keep the amount of dye used, the place you're distributing to, the location, temp... the list goes on. Your dependent variable is the outcome: Which food dye diffuses first.

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A 73.9 kg weight-watcher wishes to climb a

Tresset [83]

The height to which the weight-watcher must climb to work off the equivalent 991 (food) Calories is 0.59 Km

<h3>How to determine the energy. </h3>

1 food calorie = 103 calories

Therefore,

991 food calories = 991 × 103

991 food calories = 102073 calories

Multiply by 4.2 to express in joule (J)

991 food calories = 102073 × 4.2

991 food calories = 428706.6 J

<h3>How to determine the height </h3>

Energy (E) = 428706.6 J
Mass (m) = 73.9 kg
Acceleration due to gravity (g) = 9.8 m/s²

Height (h) =?

E = mgh

Divide both side by mg

h = E / mg

h = 428706.6 / (73.9 × 9.8)

h = 591.95 m

Divide by 1000 to express in km

h = 591.95 / 1000

h = 0.59 Km

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Which type of compound is salt

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3 years ago

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find the gravitational force between the two bodies of unit mass each and separated by unit distance

Ede4ka [16]

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Here gravitational force is G or gravity constant.

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4 years ago

A man is pulling a 13-kg sled across a flat, snowy surface. He holds the handle of the sled at a 30° angle with the ground.

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4 years ago

A 4.3 kg steel ball and 6.5 m cord of negligible mass make up a simple pendulum that can pivot without friction about the point

Alekssandra [29.7K]

Answer

given,

mass of steel ball, M = 4.3 kg

length of the chord, L = 6.5 m

mass of the block, m = 4.3 Kg

coefficient of friction, μ = 0.9

acceleration due to gravity, g = 9.81 m/s²

here the potential energy of the bob is converted into kinetic energy

$m g L = \dfrac{1}{2} mv^2$

$v= \sqrt{2gL}$

$v= \sqrt{2\times 9.8\times 6.5}$

v = 11.29 m/s

As the collision is elastic the velocity of the block is same as that of bob.

now,

work done by the friction force = kinetic energy of the block

$f . d = \dfrac{1}{2} mv^2$

$\mu m g. d = \dfrac{1}{2} mv^2$

$d=\dfrac{v^2}{2\mu g}$

$d=\dfrac{11.29^2}{2\times 0.9 \times 9.8}$

d = 7.23 m

the distance traveled by the block will be equal to 7.23 m.

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3 years ago