A force is described by its BLANK and by the direction in which it acts

A rope is vibrated so that transverse waves propagate down it. if the distance between crests is 0.5 m and a new crest reaches t

xxMikexx [17]

The speed of a wave, V, is f *lambda. Where f is the frequency and lambda is the distance. If a new crest reaches the end every 4 secs; it takes 8s to cover the distance. Hence, f, which is the number of oscillations covered is 8s. So we have V = 8 * 5 = 40 ms^1.

7 0

3 years ago

Easy chemical formula question (9th grade)

kogti [31]

Silver and Silver (II) are different because silver (II) has more than 1 electric charge. It has 2! lol

4 0

3 years ago

Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.

Gemiola [76]

Answer:

a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

V = i R

i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

R = 4 Ω

i₄ = V / 4

R = 8 Ω

i₈ = V / 8

we look for the relationship between these two currents

i₈ /i₄ = 4/8 = ½

i₈ = 0.5 i₄

R = 3 Ω

i₃ = V3

R = 10 Ω

i₁₀ = V / 10

we look for relationships

i₁₀ / 1₃ = 3/10

i₁₀ = 0.3 i₃

i₁₀ / 1₈ = 8/10

i₁₀ = 0.8 i₈

7 0

2 years ago

PLZ HELP I WILL GIVE BRAINLIEST

Vaselesa [24]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of diver = 77kg

Height of jump = 8.18m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we apply the motion equation below:

v² = u² + 2gH

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

H is the height

Now insert the parameters and solve;

v² = 0² + 2 x 9.8 x 8.18

v = 12.7m/s

8 0

2 years ago

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

3 years ago