Ask question

Ask question

All categories

svet-max [94.6K]

3 years ago

5

About when was the sail developed, making travel possible on the open water?

1 answer:

dalvyx [7]3 years ago

4 0

The first sail was developed in 1300 BC in Mesopotamia

You might be interested in

Which are the products in the equation CH3SH + 4O2 → CO2 + SO2 +2H2O? Check all that apply. CH3SH, O2 ,CO2, SO2, H2O

VikaD [51]

Answer:

C D E is the answer

Explanation:

7 0

3 years ago

Read 2 more answers

Which rock is made mostly of dark, fine-grained silicate minerals, chiefly plagioclase feldspar and pyroxene, and magnetite?

Dafna1 [17]

A dark-colored (igneous rock,) commonly extrusive from volcanic eruptions and composed primarily of the minerals of calcic plagioclase and pyroxene, and sometimes olivine. Basalt is the fine-grained equivalent of gabbro.

8 0

4 years ago

n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

The potential of the soap bubble is $V_1 =307.0 V$

The new radius of the soap bubble is $R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m$

The initial electric potential is mathematically represented as

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

So

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

where k is the coulomb's constant with value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

8 0

3 years ago

All living and nonliving material is composed of<br> (1) air (3) water<br> (2) elements (4) soil

Elan Coil [88]

Elements is the answer to this question

5 0

3 years ago

Read 2 more answers

A weightlifter curls a 30 kg bar, raising it each time a distance of 0.50 m .How many times must he repeat this exercise to burn

ludmilkaskok [199]

Answer:

$N = 2141 times$

Explanation:

Each time the work done to raise a given mass is

$W = mgh$

here we know that

$m = 30 kg$

$h = 0.50 m$

now we have

$W = (30 kg)(9.81 m/s^2)(0.50)$

$W = 147.15 J$

since it is just 25% of actual energy consumed as we know its efficiency is 25%

so we have total energy consumed in this way

$E_{total} = \frac{147.15}{0.25}$

$E_{total} = 588.6 J$

now if it took N number of times so burn the fat of a pizza then

$N(588.6) = 1260 \times 10^3$

$N = 2141 times$

4 0

3 years ago