All categories

4 years ago

Find the exponent A in the equation

Physics

1 answer:

hodyreva [135]4 years ago

5 0

Start by swapping the variables in the equations with their units (leave A alone for now): $v=\frac{a^2t^A}{x} \Rightarrow \frac{L}{T} = \frac{ (\frac{L}{T^2})^2 T^A }{L}$

Multiply by L and square the fraction: $\frac{L^2}{T} = \frac{L^2}{T^4} T^A \Rightarrow \frac{L^2}{T} \div \frac{L^2}{T^4}=T^A$

Divide the fraction by multiplying by the reciprocal: $\frac{L^2}{T} \times \frac{T^4}{L^2} = T^A \Rightarrow T^3=T^A$

Therefore A=3

You might be interested in

64 g of sulfur dioxide (SO2) contains 32 g of oxygen. Calculate how much sulfur it contains.

Ghella [55]

In sulfur dioxide, there are 2 oxygen atoms and 1 sulfur atom. As there are 32g of sulfur and 32g of oxygen, that would mean that each oxygen atom would weigh about 16g. Given that, the mass of a single sulfur atom is twice that of a single oxygen atom.

5 0

3 years ago

Read 2 more answers

The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal re

inna [77]

Answer:

Explanation:

The thermal efficiency of a Power cycle $\eta = \dfrac{Q_H -Q_c}{Q_H}$

where;

$\eta = 50\% = 0.5$

$Q_H = Heat \ flow \ from \ higher \ temperature$

$Q_c = Heat \ flow \ from \ lower \ temperature$

$0.5 = \dfrac{Q_H -Q_c}{Q_H}$

$0.5 Q_H = Q_H - Q_c$ --- (1)

$Q_c = 0.5 Q_H$ ---- (2)

The coefficient of performance is:

$COP_R = \dfrac{Q_c}{Q_H -Q_c}$

let replace the value of $Q_c = 0.5 Q_H$ in the above equation then;

$COP_R = \dfrac{0.5Q_H}{Q_H -0.5 Q_H}$

$COP_R = \dfrac{0.5Q_H}{0.5 Q_H}$

$COP_R = 1$

The

On the other hand, the heat pump

$COP_{HP} = \dfrac{Q_H}{Q_H -Q_c}$

By replacing equation (1) into the above equation; we have:

$COP_{HP} = \dfrac{Q_H}{0.5Q_{H}}$

$COP_{HP} = \dfrac{1}{0.5}$

$COP_{HP} =2$

5 0

3 years ago

When the moon slowly moves away from the earth.How is it affecting the gravitational force between the earth and the moon?

sukhopar [10]

Well, think about how the tides will be affected when the moon moves farther away. If the moon first started off very close the earth, we would have more tsunamis. (Scientists have found that the moon has possibly been closer to earth long ago.) While it moves away, soon there will no longer be many tides.

3 0

3 years ago

A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin

Greeley [361]

Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Electric Resistance</u>

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

$\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

$\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}$

This sum is repeated 10 times. Operating each term:

$\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}$

All the terms have the same denominator, thus:

$\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}$

Taking the reciprocals:

$R_e=R/100$

The equivalent resistance of the combination is R/100

6 0

3 years ago

A sound wave has a frequency of 425 Hz. What is the period of this wave

Oduvanchick [21]

Answer: The time period of the given wave with frequency of 425 Hertz is 0.0023 seconds.

Explanation:

Frequency of the wave = 425 Hertz = $425 sec^{-1}$

$\text{Time period}=\frac{1}{Frequency}$

$\text{Time period}==\frac{1}{425 sec^{-1}}=0.0023 sec$

The time period of the given wave with frequency of 425 Hertz is 0.0023 seconds.

5 0

3 years ago