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3 years ago

What happens when there is fluvial erosion?

1 answer:

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Froghopper insects have a typical mass of around 11.3 mg and can jump to a height of 58.8 cm. The takeoff velocity is achieved a

allochka39001 [22]

Answer:

2874.33 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 0.588 - 0.002 = 0.586 m

The final velocity will be the initial velocity

$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.586\\\Rightarrow a=-\frac{0.586\times -9.81}{0.002}\\\Rightarrow a=2874.33\ m/s^2$

Acceleration of the frog is 2874.33 m/s²

6 0

3 years ago

Can someone help with this???

Rus_ich [418]

no BECQUSE POSUM BROOB SHSHSJ

8 0

3 years ago

Read 2 more answers

The balmer series is formed by electron transitions in hydrogen that

RSB [31]

Answer:

The Balmer series refers to the spectral lines of hydrogen, associated to the emission of photons when an electron in the hydrogen atom jumps from a level $n \geq 3$ to the level $n=2$ .

The wavelength associated to each spectral line of the Balmer series is given by:

$\frac{1}{\lambda}=R_H (\frac{1}{2^2}-\frac{1}{n^2})$

where $R_H$ is the Rydberg constant for hydrogen, and where $n$ is the initial level of the electron that jumps to the level n = 2.

The first few spectral lines associated to this series are withing the visible part of the electromagnetic spectrum, and their wavelengths are:

656 nm (red, corresponding to the transition $3 \rightarrow 2$ )

486 nm (green, $4 \rightarrow 2$ )

434 nm (blue, $5 \rightarrow 2$ )

410 nm (violet, $6 \rightarrow 2$ )

All the following lines lie in the ultraviolet part of the spectrum. The limit of the Balmer series, corresponding to the transition $\infty \rightarrow 2$ , is at 364.6 nm.

4 0

3 years ago

At what speed does a typical electron pass by any given point in the wire?

attashe74 [19]

A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.

a) How many electrons pass through the light bulb each second?

b) What is the current density in the wire? (answer in A/m^2)

c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)

a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s

b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²

c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³

(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed

(q/m³)(A)(v) = i

v = i.[(q/m³)A]ˉ¹

v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s

7 0

3 years ago

Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of K

Drupady [299]

Answer:

$3.60432\times 10^{26}\ kg$

Explanation:

a = Orbital radius = $1.62\times 10^8\ m$

T = Orbital period = 23.21 hours

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

From Kepler's third law we get

$M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg$

From the given data the mass of Saturn is $3.60432\times 10^{26}\ kg$

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2 years ago