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3 years ago

What is the sl unit for momentum

Physics

1 answer:

mario62 [17]3 years ago

8 0

Momentum is mass times velocity and mass is measured in kg and velocity is measured per second so it really depends.

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. En 1/2 hora, un ciclista recorrió 20 kilómetros. ¿Cuál fue el promedio de velocidad del ciclista? a. 10 km / h b. 60 km c. 40

ohaa [14]

Distancia el lingo por la tarde

8 0

2 years ago

What is the deBroglie wavelength of an electron moving at 1.37 x 106 m/s if the mass of the electron is 9.11 x 10-28 g

Nina [5.8K]

Answer:

The value is $\lambda = 5.30 *10^{-10 } \ m$

Explanation:

From the question we are told that

The velocity of the electron is $v = 1.37 *10^{6} \ m/s$

The mass of the electron is $m = 9.11 *10^{-28} \ g = 9.11 *10^{-31} \ kg$

Generally the deBroglie wavelength is mathematically represented as

$\lambda = \frac{h}{mv}$

Here h is the Planck'c constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

$\lambda = \frac{6.62607015 * 10^{-34} }{9.11 *10^{-31} * 1.37 *10^{6}}$

=> $\lambda = 5.30 *10^{-10 } \ m$

7 0

3 years ago

1. A 2.10 m rope attaches a tire to an overhanging tree limb. A girl swinging on the

musickatia [10]

Answer:

29.6 kg

Explanation:

Centripetal force = mass × centripetal acceleration

F = m v² / r

88.0 N = m (2.50 m/s)² / 2.10 m

m = 29.6 kg

3 0

2 years ago

What is the term for the pattern of spiral impressions inside a handgun?

Korolek [52]

It's called rifling, to stabilise the bullet.

4 0

3 years ago

Read 2 more answers

Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

5 0

2 years ago