Ask question

Ask question

All categories

3 years ago

5

The area of the bar over r = 2 is 0.234. what is the area of the bar over r = 4?

1 answer:

natita [175]3 years ago

4 0

The area of the bar over r=4 is 0.468

You might be interested in

Two airplanes leave an airport at the same time. The velocity of the first airplane is 750 m/h at a heading of 51.3°. The veloci

abruzzese [7]

Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.

The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
163° - 51.3° = 111.7°

Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles

Answer: 2335.4 miles

4 0

3 years ago

Read 2 more answers

Jenny pushes a 40 N crate down the hall 2m. How much work did she do?

Shalnov [3]

Work done = force x distance = 40 x 2 = 80 Joules.

8 0

2 years ago

Define force and provide an example

ollegr [7]

Answer:

force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.

4 0

3 years ago

How many atoms are in 3MgCl2?

Bogdan [553]

Answer:

3MgCl2 has 9 atoms.

Explanation:

The Element Magnesium (Mg) has 3 atoms.

The Element Chloride (Cl) has 6 attoms.

Their fore 6 + 3 is 9 of course. 3MgCl2 has 9 atoms.

BTW: 3MgCl2 is a molecular compound as well as H2O and CO2.

8 0

2 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = <em>0.85c </em>

7 0

3 years ago