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tatyana61 [14]
3 years ago
13

When a particle of mass m is at (x comma 0 )​, it is attracted toward the origin with a force whose magnitude is StartFraction y

Over x squared EndFraction . If the particle starts from rest at x equals b and is acted upon by no other​ forces, find the work done on it by the time it reaches x equals a​, 0 less than a less than b.
Physics
1 answer:
Dahasolnce [82]3 years ago
8 0

Answer:

W = y (b-a) / ab

Explanation:

Work is defined by the expression

W = ∫ F. dr

In this case the force is in the same direction of displacement, so the scalar product is reduced to the ordinary product

W = ∫ F dr

The expression of the strength left is

F = -y / x²

let's replace and integrate

W = ∫ (-y / x²) dx

W = -y (-1 / x)

We evaluate between the lower limit x = b + a to the upper limit x = 0 + a

W = -y (-1 / b + 1 / a)

W = y (b-a) / ab

where (b-a) is the distance traveled

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Alpha particles
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8 0
3 years ago
A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite
DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx  is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x  4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x  4 π

≥  . 5254 x ⁻²⁵

h / λ ≥  . 5254 x ⁻²⁵

 6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ  

12.56 x 10⁻⁹ ≥ λ  

longest wave length = 12.56 n m

6 0
3 years ago
A certain ocean wave has a frequency of 0.07 hertz and a wavelength of 10 meters. what is the wave's speed? a 0.07 m/s
Elis [28]
V=f*wavelength
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8 0
3 years ago
A car of mass 1500 kg travels due East with a constant speed of 25.0 m/s. Eventually it turns right, and travels due South with
e-lub [12.9K]

Answer:

The direction of the car’s change in linear momentum is 149.04° West of North

Explanation:

Momentum is defined as the product of mass of a body and its velocity

Momentum = mass × velocity

Change in Momentum = mass × change in velocity

∆P = m∆v

∆P = m(v-u)

Given m = 1500kg

v = 25m/s

u = 15m/s

∆P = 1500(25-15)

∆P = 1500×10

∆P = 15,000kgm/s

Since the car first travels due East i.e +x direction

x = 25m/s

Travelling due south is negative y direction

y = -15m/s

Direction of the car change

θ = tan^-1(y/x)

θ = tan^-1(-15/25)

θ = tan^-1(-0.6)

θ = -30.96°

Since tan is negative in the second quadrant

θ = 180-30.96

θ = 149.04°

The direction of the car’s change in linear momentum is 149.04° West of North

5 0
4 years ago
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