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3 years ago

Which statements describe Rutherford model of the atom

Physics

2 answers:

il63 [147K]3 years ago

4 0

Answer:

The nucleus is small and Very delicate. The electrons, for its turn it orbits around your nucleus. As the atom is the control/union of nucleus and electrons, and other particles, it is fair to say that an atom is a mostly empty space

Explanation:

Goshia [24]3 years ago

3 0

Answer:

rutherford had the "plum pudding model"

Explanation:

so basically its a circle filled with lots of electrons (negativly charged particles)

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What is the acceleration of the object? ______ m/s2

professor190 [17]

Answer:

Explanation:

this graph show the decrease in velocity with time

at time t=0 velocity is 40 m/sec

at time t=12sec velocity is decreases from 40 to 10 m/s

so the acceleration of the object at time=12 sec is

a=v/t

a=Δv/t

Δv=v1-v2=40-10=30 m/s

a=30/12=2.5 m/s^2

8 0

4 years ago

A sodium surface is illuminated with light of wavelenght 300nm. The work function of the metal is 2.4eV.

denis-greek [22]

Answer:

1) 1.67eV

2) 505nm

Explanation:

The maximum kinetic energy of photoelectrons ,

KEmax=λhc−W=(0.3×10−6)(1.6×10−19)(6.62×10−34)(3×108)eV−2.46eV=1.67eV

If λ0 is the cut-off wavelength, W=λ0hc

or λ0=Whc=2.46×1.6×10−19(6.62×10−34)(3×108)=5.05×10−7m=505×10−9=505nm

We know that the work function is the minimum photon energy for taking place of photoelectric effect.

5 0

3 years ago

Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field

kirza4 [7]

Answer: the work done by the force is 0

Explanation:

F (x², xy)

121 = 11²

so R = x² + y² = 11²

p = x². Q = xy

Δp/Δy = 0, ΔQ/Δx

using Green's theorem

woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA

= (x² + y² = 121)_∫∫ yΔA

now let x = rcosФ, y = rsinФ

ΔA = rΔrΔФ

so r from 0 to 11

and Ф from 0 to 2π

= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ

= 0_∫^2π SinФΔФ 0_∫^11 r²Δr

= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0

therefore the work done by the force is 0

3 0

4 years ago

A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?

madam [21]

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy $=\frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

3 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago