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Natasha_Volkova [10]
3 years ago
11

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles trave

led in a straight line and some were deflected at different angles.
Physics
1 answer:
jeyben [28]3 years ago
4 0

The question is incomplete, the complete question is;

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles traveled in a straight line and some were deflected at different angles. Which statement best describes what Rutherford concluded from the motion of the particles? Some particles traveled through empty spaces between atoms and some particles were deflected by electrons. Some particles traveled through empty parts of the atom and some particles were deflected by electrons. Some particles traveled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms. Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Explanation:

Rutherford first proposed the nuclear model of the atom after his landmark experiment.

In this experiment, alpha particles from a source was focused on a thin gold foil. Some of the particles passed through empty spaces within the atom but were deflected at different angles by a small area of high-density positive charge within an atom which Rutherford later called the atomic nucleus.

Hence the answer above.

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A stone is tied to a 0.85-meter cord. it is swung in a circle at a constant rate of 6.0 m/s. what is the centripetal acceleratio
Andrei [34K]

Answer:42.4m/s^2

Explanation:

Velocity(v)=6m/s

Radius(r)=0.85 meter

Centripetal acceleration=(v x v) ➗ r

Centripetal acceleration=(6 x 6) ➗ 0.85

Centripetal acceleration=36 ➗ 0.85

Centripetal acceleration=42.4

5 0
3 years ago
A net force of 100 newtons is applied to a wagon for 5 seconds. This causes the wagon to undergo a change in momentum of
Nataly_w [17]
If it was applied for 5 seconds for 100 Newtons, than the change in momentum is 100*5=20 kg m/s
4 0
4 years ago
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 41.0 lb/in2. If the pump is a cylinder of lengt
damaskus [11]

Answer:

L - h = 12.3672 in

Explanation:

Given

P = 41.0 lb/in² = 41 P.S.I

L = 16.8 in

A = 3.00 in²

h = ?

In order that air flows into the tire, the pressure in the pump must be more than the tire pressure,  41.0  PSI.

We assume that air follows ideal gas equation, the temperature of the compressed air remains constant as the piston moves down. Taking one atmospheric pressure to be  14.6959  P.S.I , we can use the ideal gas equation

P*V = n*R*T

As number of moles of air do not change during its compression in the pump, n*R*T of the gas equation is constant. Therefore we have

P₁*V₁ = P₂*V₂    ⇒    V₂ = P₁*V₁ / P₂

where  

1  and  2  are initial and final states respectively,

V₁ = A*L = (3.00 in²)*(16.8 in)   ⇒   V₁ = 50.4 in³

P₁ = 14.6959  P.S.I

P₂ = P₁ + P = (14.6959 lb/in²) + (41.0 lb/in²) = 55.6959 lb/in²

Inserting various values we get

V₂ = (14.6959  P.S.I)*(50.4 in³) / (55.6959 lb/in²)

⇒  V₂ = 13.2985 in³

Length of pump, measured from bottom, this volume corresponds to is

h = V₂ / A  = (13.2985 in³) / (3.00 in²)

⇒  h = 4.4328 in

Piston must be pushed down by more than

L - h = 16.8 in - 4.4328 in = 12.3672 in

4 0
3 years ago
On Earth, all objects dropped in freefall will:
masya89 [10]

Answer:

it will have a speed of 10ms after 1 second

8 0
3 years ago
The person has an upward acceleration of 0.70 m/s^2 and is lifted from rest through a distance of 13 m. A rescue helicopter lift
Alex

Answer:

a)

840 N

b)

10920 J

c)

- 10192 J

d)

4.3 m/s

Explanation:

a)

T = tension force in the cable in upward direction = ?

a = acceleration of the person in upward direction = 0.70 m/s²

m = mass of the person being lifted = 80 kg

Force equation for the motion of person in upward direction is given as

T - mg = ma

T = m (g + a)

T = (80) (9.8 + 0.70)

T = 840 N

b)

d = distance traveled in upward direction = 13 m

W_{t} = Work done by tension force

Work done by tension force is given as

W_{t} = T d

W_{t} = (840) (13)

W_{t} = 10920 J

c)

d = distance traveled in upward direction = 13 m

W_{g} = Work done by person's weight

Work done by person's weight is given as

W_{g} = - mg d

W_{g} = - (80 x 9.8) (13)

W_{g} = - 10192 J

d)

F_{net} = Net force on the person = ma = 80 x 0.70 = 56 N

v₀ = initial speed of the person = 0 m/s

v = final speed

Using work-energy theorem

F_{net} d = (0.5) m (v² - v₀²)

(56) (13) = (0.5) (80) (v² - 0²)

v = 4.3 m/s

6 0
4 years ago
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