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Novay_Z [31]
3 years ago
9

In his famous experiment, Rutherford fired alpha particles at a thin gold film. Most of the alpha particles went through the fir

m and a very few bounced back. Suppose instead that about one-half the alpha particles bounced back and one-half went through. How would this have changed his conclusion about the structure of the atom?
1. The nucleus is about 1/2 the size of the atom.
2. The atom has no nucleus.
3. None of these
4. The nucleus covers the entire atom.
5. The atom is mostly empty space.
Physics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

1. The nucleus is about 1/2 the size of the atom

Explanation:

Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size

Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.

Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles

so correct answer will be

1. The nucleus is about 1/2 the size of the atom

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The nervous system has two distinct branches. They are the:
Troyanec [42]

Answer:

central nervous system

peripheral nervous system

Explanation:

The nervous system has two main parts: The central nervous system is made up of the brain and spinal cord. The peripheral nervous system is made up of nerves that branch off from the spinal cord and extend to all parts of the body.Oct 1, 2018

3 0
3 years ago
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A steel marble with 0.05 kg of mass starts from rest and rolls down a ramp. It travels 0.25 m in 1.2 seconds. Find the force act
emmasim [6.3K]

After finding acceleration, it is found that 0.02 N of force is acting on the marble

<h3>What is Force ?</h3>

Force can simply be defined as a pull or push. It is the product of mass and acceleration of the object. It is a vector quantity and it is measured in Newton.

Given that a steel marble with 0.05 kg of mass starts from rest and rolls down a ramp. It travels 0.25 m in 1.2 seconds.

The parameters to consider are;

  • m = 0.05 Kg
  • u = 0
  • s = 0.25 m
  • t = 1.2 s
  • a = ?
  • F = ?

Before we find the force acting on the marble, let us first find the acceleration by using the formula: s = ut + 1/2at²

Substitute all the parameters into the formula

0.25 = 0 + 1/2 × a × 1.2²

0.25 = 1/2 × a × 1.44

0.25 = 0.72a

a = 0.25/0.72

a = 0.35 m/s²

The force acting on the marble will be ;

F = ma

F = 0.05 × 0.35

F = 0.017

F = 0.02 N

Therefore, the force acting on the marble is 0.02 N

Learn more about Force here: brainly.com/question/388851

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3 0
1 year ago
What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)
Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity g=10m/sec^2

And final value of acceleration due to gravity g=9.8m/sec^2

We have to find the percentage error

We know that percentage error is given by percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100

So percentage\ error=\frac{10-9.8}{10}\times 100=2 %

4 0
3 years ago
A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F
EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B =  1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ)   eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0
3 years ago
Please help me with this question
snow_lady [41]

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