How does Mr. Anderson define solubility in the video?

The Carson family's pancake recipe uses 2 teaspoons of baking powder for every 1/3 of a teaspoon of salt. How much baking powder

melomori [17]

Answer:

6 teaspoons of baking powder required.

Explanation:

Given that

According to the recipe of pancake,

For every $\frac{1}{3}$ teaspoon of salt, 2 teaspoons of baking baking powder is required.

To find:

How much baking powder will be needed, if 1 teaspoon of salt was used ?

Solution:

This problem can be solved using ratio.

$\frac{1}3$ teaspoon of salt : 2 teaspoons of baking powder

Let us multiply the above ratio with 3.

$\frac{1}{3}\times 3$ teaspoon of salt : 2 $\times$ 3 teaspoons of baking powder

OR

1 teaspoon of salt : 6 teaspoons of baking powder

So, answer is 6 teaspoons of baking powder required.

Also, we can use the unitary method:

$\frac{1}3$ teaspoon of salt needs = 2 teaspoons of baking powder

1 teaspoon of salt needs = $\frac{2}{\frac{1}3}$ teaspoons of baking powder

1 teaspoon of salt needs = 2 $\times$ 3 = 6 teaspoons of baking powder needed

So, the answer is:

6 teaspoons of baking powder required.

5 0

3 years ago

Read 2 more answers

In designing a backyard water fountain, a gardener wants to stream of water to exit from the bottom of one tub and land in a sec

mote1985 [20]

To solve this problem it is necessary to apply the concepts related to the kinematic equations of movement description.

From the definition we know that the speed of a body can be described as a function of gravity and height

$V = \sqrt{2gh}$

$V = \sqrt{2*9.8*0.15}$

$V = 1.714m/s$

Then applying the kinematic equation of displacement, the height can be written as

$H = \frac{1}{2}gt^2$

Re-arrange to find t,

$t = \sqrt{2\frac{h}{g}}$

$t = \sqrt{2\frac{0.5}{9.8}}$

$t = 0.3194s$

Thus the calculation of the displacement would be subject to

$x = vt$

$x =1.714*0.3194$

$x = 0.547m$

Therefore the required distance must be 0.547m

4 0

3 years ago

A student left a bar of chocolate in the sun on a hot day. As the chocolate melted, which property changed?

barxatty [35]

The property changed was its weight

The melted chocolate weighs the same, has an identical mass, and is still composed of chocolate.

Now it’s just melted chocolate instead of a candy bar and has thus changed shapes.

7 0

3 years ago

Glucose is the main source of energy. True or false

Eddi Din [679]

Answer:

True :p

Explanation:

Carbohydrates are foods that get converted into glucose.

5 0

3 years ago

An Apple with the mass of 200g falls from the tree.what is the acceleration of the apple towards the earth.what is the accelerat

notsponge [240]

Answer:

Assume that the earth is a sphere (with radius equal to its actual radius at the equator) of uniform density.

Acceleration of the apple towards the earth: approximately $9.79\; \rm m \cdot s^{-2}$ .

Acceleration of the earth towards the apple: approximately $3.28 \times 10^{-25}\; \rm m \cdot s^{-2}$ .

Both values were rounded to three significant figures. Air resistance is assumed to be negligible.

Explanation:

Convert the mass of the apple to standard units:

$m(\text{apple}) = 200\; \rm g = 0.200\; \rm kg$ .

Look up the mass and radius of the earth:

Mass of the earth: $m(\text{earth})\approx 5.97\times 10^{24}\; \rm kg$ .
Radius of the earth: $r \approx 6.3781\times 10^{6}\; \rm m$ (at the equator.)

Assume that the earth is a sphere of uniform density. The acceleration of an object moving towards the earth under free fall would be approximately equal to the gravitational field strength of the earth at that position.

For a sphere of mass $m$ (assuming uniform density,) the gravitational field strength $g$ at a distance of $r$ away from the center of the sphere would be:

$\displaystyle g = \frac{G \cdot m}{r^2}$ .

The height of the tree should be much smaller than the radius of the earth. Therefore, the distance between the apple and the center of the earth would be approximately equal to the radius of the earth.

The strength of the gravitational field of the earth at the position of the apple would be approximately:

$\begin{aligned} & \frac{G \cdot m(\text{earth})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 5.97 \times 10^{24}\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 9.79\; \rm m\cdot s^{-2} \end{aligned}$ .

This value should be approximately equal to the acceleration of the apple towards the earth.

On the other hand, assume that the apple acts like a point mass when compared to the earth. In other words, assume that the radius of the apple is much smaller than the distance between the apple and the earth.

It can be shown (using calculus) that if the earth is a sphere with uniform density, the earth will behave just like a point mass at the center of the earth when studying interactions between the earth and objects outside of it.

At the center of the earth, the strength of the gravitational field due to the apple would be:

$\begin{aligned} & \frac{G \cdot m(\text{apple})}{r^2} \\ &\approx \frac{6.67\times 10^{-11}\; \rm m^3\cdot kg^{-1}\cdot s^{-2} \times 0.200\; \rm kg}{\left(6.3781\times 10^{6}\; \rm m\right)^2} \\ &\approx 3.28 \times 10^{-25}\; \rm m\cdot s^{-2} \end{aligned}$ .

Under the assumption that the earth is a sphere of uniform density, this value should be equal to the acceleration of the earth towards the apple due to the gravitational attraction of the apple.

6 0

3 years ago