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3 years ago

9- Where did Rutherford propose that most of the mass of an atom

Physics

1 answer:

siniylev [52]3 years ago

8 0

Answer:

A) in the middle of the atom

Explanation:

He concluded that all of the positive charge and the majority of the mass of the atom must be concentrated in a very small space in the atom's interior, which he called the nucleus.

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A smooth circular hoop with a radius of 0.400 m is placed flat on the floor. A 0.325-kg particle slides around the inside edge o

aleksandrvk [35]

Answer:

a) W = - 6.825 J, b) θ = 1.72 revolution

Explanation:

a) In this exercise the work of the friction force is negative and is equal to the variation of the kinetic energy of the particle

W = ΔK

W = K_f - K₀

W = ½ m v_f² - ½ m v₀²

W = ½ 0.325 (5.5² - 8.5²)

W = - 6.825 J

b) find us the coefficient of friction

Let's use Newton's second law

fr = μ N

y-axis (vertical) N-W = 0

fr = μ W

work is defined by

W = F d

the distance traveled in a revolution is

d₀ = 2π r

W = μ mg d₀ = -6.825

μ = $\frac{ -6.825}{d_o \ mg}$

The total work as the object stops the final velocity is zero v_f = 0

W = 0 - ½ m v₀²

W = - ½ 0.325 8.5²

W = - 11.74 J

μ mg d = -11.74

we subtitle the friction coefficient value

( $\frac{-6.8525 }{d_o mg}$ ) m g d = -11.74

6.825 $\frac{d}{d_o}$ = 11.74

d = 11.74/6.825 d₀

d = 1.7201 2π 0.400

d = 4.32 m

this is the total distance traveled, the distance and the angle are related

θ = d / r

θ = 4.32 / 0.40

θ = 10.808 rad

we reduce to revolutions

θ = 10.808 rad (1rev / 2π rad)

θ = 1.72 revolution

3 0

3 years ago

Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo

LenaWriter [7]

Answer:

$6.33\times 10^8\ kg\cdot m/s$

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = $33\times \frac{5}{18}=9.17\ m/s$

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

= $(6.9\times 10^7\ kg)(9.17\ m/s)$

= $6.33\times 10^8\ kg\cdot m/s$

Therefore, the magnitude of ship's momentum is $6.33\times 10^8\ kg\cdot m/s$ .

6 0

4 years ago

If neutron stars are squeezed harder they collapse into black hole; how would this transition occur?

antoniya [11.8K]

Answer:

If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.

Explanation:

hope this helps have a good night :)

6 0

3 years ago

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta

bonufazy [111]

Answer:

acceleration = 22.14 m/s²

time = 2.80 s

Explanation:

given data

initial velocity = 31.0

final velocity =

time = 1.40

solution

we get here acceleration that is

acceleration = $\frac{v(final) - v(initial)}{t}$ ...............1

acceleration = $\frac{31-0}{1.40}$

acceleration = 22.14 m/s²

and

by kinematic equation of motion

v = u +at .........2

t = $\frac{62-0}{22.14}$

time = 2.80 s

8 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

3 0

3 years ago

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