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3 years ago

Dfine ' weight of moist air '

Physics

1 answer:

grandymaker [24]3 years ago

6 0

Answer:

First of all, “moist air” is air with a high water vapor content. Water vapor, the invisible, gaseous form of water, occurs in highly variable amounts in the atmosphere. Water is composed of a hydrogen atom and two oxygen atoms (H2O) and has a molecular weight of 18 grams per mole.

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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

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

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$

Try dealing with the powers of 10 first: On the right, we have

$\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}$

Meanwhile, the other values on the right reduce to

$\dfrac{6.67\times5.98}{3.8^2}\approx2.76$

Then taking units into account, we end up with the equation

$2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2$

Now we solve for $m_2$ :

$m_2=\dfrac{2.0\times10^{20}\,\mathrm N}{2.76\times10^{-3}\,\frac{\mathrm N}{\mathrm{kg}}}\approx0.725\times10^{20-(-3)}\,\mathrm{kg}$

$m_2=7.25\times10^{22}\,\mathrm{kg}$

or, if taking significant digits into account,

$m_2=7.3\times10^{22}\,\mathrm{kg}$

5 0

3 years ago

In which direction does energy flow in an ecosystem

muminat

QUESTION -

In which direction does energy flow in an ecosystem ?

ANSWER -

The flow of energy in an ecosystem is best described as energy moving in one direction from the sun to the producers then to the consumers. Explanation; Energy flow is the amount of energy that moves through successive trophic levels of a food chain in an ecosystem

4 0

3 years ago

Read 2 more answers

A pitcher throws a 0.140 kg baseball, and it approaches the bat at a speed of 35.0 m/s. The bat does Wnc = 75.0 J of work on the

Eva8 [605]

Answer:

The speed of the ball is 42.5 m/s

Explanation:

The initial kinetic energy of the ball is:

$K_1=\frac{1}{2} m v_0^2=\frac{1}{2}*0.140 kg*(35.0 m/s)^2$ = 85.75 J

The speed of the ball after leaving the bat is:

$K_2=K_1+W_{nc}\\ \frac{1}{2}mV^2= 85.75 J + 75 J\\ (\frac{1}{2}mV^2)2=( 160.75 J)2\\ mV^2= 321.5 J\\ V^2= \frac{321.5 J}{0.140kg} \\ V=\sqrt{\frac{321.5 J}{0.140kg}}$